Aits Part Test - Ii_qn & Sol

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ALL INDIA TEST SERIESJEE(Advanced)2013PART TEST - II (PAPER - 1 amp 2 )

(9th December 2012)

QUESTIONS ANSWERS amp SOLUTIONS )

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PART -I

ANSWERS HINTS amp SOLUTIONSPART TEST - II

PAPER-1

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Physics

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(A) -) (q r 5) (A) -) (p) (A) -) (q r s)(B) -) (r) (B) -) (p q) I (8) -~ (r)(C) -4 (q) (C) -) (5) (C) -) (q)(0) -) (p) (0) -) (r) (O) -)(p)

(A) -) (p) (A) -) (q rs) (A) -) (r)(B) -) (p q) (B) -) (r) (B) -) (r)(Cl -) (5) (C) -) (q) (C) -) (p)(0) -) (r) (0) -) (p) (0) -) (q)

(A) -) (r) (A) -) (r) (A) -) (p)(B)-) (r) (B) -) (r) (B) -) (p q) I(C) -) (p) (C) -) (p) (C) -4 (5) ~(0) -) (q) (0) -~ (q) (0) -) (r)

(A) -) (r)

(B) -) (r)

(C) -) (p)

(0) -) (q)

(A) -) (p)

(B) -) (p q)(C) -) (s)(0) -) (r)

(A) -)(q f s)(B) -) (r)(C) -) (q)

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Mathematics PART - III

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Physics

PART -I

AIT5-PT -1HPaper-1 )-PCM(S)-JEE(Advanced)13laquo trtraquo-

SECTION-A

Single Correct Choice Type

This section contains 9 multiple choice questions Each question has four choices (A) (B) (C) and (0)out of which ONLY ONE is correct

1 In a certain region of space a unifonn and constant electric field and a magnetic field parallel toeach other are present A proton is fired from a point A in the field with speed v = 4 X 104 mls atan angle of a with the field direction The proton reaches a point B in the field where its velocity

makes an angle 13with the field direction If s~na=J3 Find the electric potential differencebull SIO 13

between the points A and B Take mp (mass of proton) = 16 x 1027 kg and e (magnitude ofelectronic charge) = 16 x 1019 C(A) 16 V (B) 163 V(C) 90 V (0) 30 V

Sol AWork done by magnetic force is zero

1 2 1 2s2 from work energy theorem - mpYB =- mpVA + q8 V 2j 2) and simultaneously there is no change of velocity component along the direction of perpendicular tqelectric and magnetic field gt

v~ sina =VB sinl3After solving w = 16 Volt

2 A metallic disc of radius r is made of a material of negligible Rresistance and can rotate about a conducting horizontalshaft A smaller non conducting disc of radius a is fixedonto the same shaft and has a massless cord wrappedaround it which is attached to a small object of mass m asshown Two ends of a resistor of resistance Rareconnected to the perimeter of the disc and to the shaft bysliding contacts The system is then placed into a unifonnhorizontal magnetic field B and the mass m is releasedFind the tenninal angular velocity with which the disc willrotate finally

1 (Take r = 10cm a = 2cm R =100 Q B = 02 T m = 50 gm g = 10 ms)

(A) 200 rads (B) 300 radls(C) 100 rads (0) 10 radls

Sol CAt tenninal stage torque applied on the smaller disc by the rope = mga

current to the disc = Bwr (where (D is terminal angular velocity)2R

B2 4torque applied by magnetic field = _Wf

4R~ 4B(Jr

So -- =mga 4RID == 100 radsec

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(D) 2n x 10-6sec

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~~hti~~~ ~I~~~~al~e~al~~~~~~~i~~dp~a~~r~~i~ ~e~ 51~r ~distance from each other and they are connected by a coil ofinductance L = 9mH as shown in figure One of the spheres (say A) is imparted an initial chargeand the01her is kept uncharged The switch S is closed at t = O After what minimum time t doesthe charge on the first sphere decrease to half of its initial value

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

BAngular frequency (f) of the LC oscillation = ~21TpoundORLrequired time is one fourth of the time period

A non conducting infinite rod is placed along the zcaxis theupper half of the rod (lying along z 2 0) is charged positivelywith a unifonn linear charge density +1while the lower half (z lt0) is charged negatively with a unifonn linear charge density -i The origin is located at the junction of the positive andnegative halves of the rod A unifonnlycharged annular disc(surface charge density aD) ofinner radius R and outer radius2R is placed in the x-y plane with its centre at the origin Findthe force on the rod due to the disc(A) 2aoR

(C) aoRCo

CThe force on the rod due to the annular disc is equal and opposite to that on the disc due to therod We take an annular strip of radius r and width dr and find the force acting on it

2 d) i IGO ddF = (ao 1tr r -- = - r2mor Co

The force2R

F = ao J dr = aoiR co Co- R

A insulating cylindrical rod of diameter d and length f ( f raquod) hasa unifonn surface charge density such that the electric field justoutside the curved surface of the cylinder at point M is Eo Find theelectlic field due to charge distribution at point P (rraquo 0WE~ ~El~

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

2 -E--- 021tlodi = modEo

S E _ todla p--- 4r2

M((rlaquoHO lt )(

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6 Electric field given by the vector E =0 ~Q(xi + Yl) NC is present in the(

x-y plane A small ring of mass M carrying charge +Q which can slidefreely on a smooth non conducting rod is projected along the rod fromthe point (0 pound) such that it can reach the other end of the rod

Assuming there is no gravity in the region What minimum velocity

should be given to the ring (in mts) jf in S1 unit QEo( =8M

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

For the electric field V=- _0_ - ~ + K2pound 2poundQE pound2 1 QE f2

from energy conservation - __ 0 - + k + _Mv2 =- - _0 - x 2 + KR 2 2 f 8

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

One mole of an ideal monoatomicgas under goes a process V = aT2 The temperature of the gas

increasesby 100 K Which of the following is incorrect(A) the work done by the gas in the process is 200 R Joule

(B) the molar heat capacity for the process is 7R 2

(C) for the process PT2 = constant

(D) for the process coetftcient of volume expansion is j ~T

For the process P = -aT

So work done by the gas = 200 R Joule 3 RT 7R

For the process molar heat capacity C = -R + -2aT = --2 V 2

I ffi 1 dV 2Vo ume expansIon coe rClent = - -- = - V dT T

A point source of heat of power P is placed at the centre of a spherical shell of mean radius Rthe material of the shell has thermal conductivity K If the temperature difference between theouterand the inner surface of the shell is not to exceed T then the thickness of the shell should not exceed

(A) 2nR2KT

(C) nR2KTP

BAt equilibrium energy radiated by point source = heat conducted through the thickness of theshell

Two infinite sheets carrying charge density +0 and -(J are arranged +XO -lt)perpendicularly as shown in the figure Select the approximate electricfield lines for the charge system near the intersection

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(B) +0

(C) (D)

Sot BTangent on the electric field lines give the direction of net electric field at that point

Assertion -Reasoning TypeI

This $ection contains 4 questions nmnbered 10 to 13 Each question contains STATEMENT-1I i

(A~sertiOn) and STATEMENT-2 (Reason) Each question has 4 choices (A) (B) (C) and (D) out of

which QNlYONE iscorrect i -----ty

10 A current of one ampere flows through a wire made of copper and aluminium as shown in figureThey have the same crosssection ASTATEMENT-1 Some charge accumulates at the boundary between the metalsandSTATEMENT2 1A

According to Gausss law ~ E (is = qenclosed for a closed surface Co

(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation forStatement-1

(B) Statement-1 is True Statement-2 is True Statement-2 is NOT a correct explanation forStatement-1

(C) Statement-lis True Statement -2 is False(D) Statement-1 is False Statement-2 is True

Sol AElectric field inside copper and aluminium are different due to different resistivity So fromgausss law the interface should contain some charge

11 1 mole of an ideal gas is contained in a cubical volume VABCDEFG at 300K One face of the cube EFGH is made up of amaterial which totally absorbs any gas molecule incident on it Atany given time

STATEMENT-1 The pressure of EFGH would be half that on ABeDand

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STATEMENT-2Gas molecules colliding with the wall EFGHare loosing ijs kinetic energy(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

Statement-1(B) Statement-1 is True Statement-2 is True Statement-2 is NOT a correct explanation for

Statement-1(C) Statement-1 is True Statement -2 is False(D) Statement-1 is False Statement-2 is True

12 In the figure shown a charged capacitor is being discharged by anexternal load resistance RlSTATEMENT-1A magnetic field is not present at point P which is situated inside thecapacitor as shown in the figure

andSTATEMENT-2From amperes law tSd7 = fLo(ienclosed) An ampere loop (circularin shape and parallelto the

plane of the plates) passing through P does not encloses any current(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

Statement-1(8) Statement-1 is True Statement-2 is True Statement-2 is NOT a correct explanation for

Statement-1(C) Statement-1 is True Statement -2 is False(D) StatEfment-1 is False Statement-2 is True

STATEMENT-1In a black body radiation energy is uniformly distributed in the radiation spectrum for allwavelengthsandSTATEMENT-2From Stefan - Boltzmanns law radiated energy from a black body is proportional to the fourthpower of its surface temperature in Kelvin(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

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Comprehension Type

This section contains 2 groups of questions Each group has 3 multiple choice question based on aparagraph Each question has 4 choices (A) (8) (C) and (D) for its answer out of which ONLY ONE iscorrect

Paragraphfor Question Nos 14 to 16

An engineer is instructed to design a tank tomJasure the heat capacity of liquids The tankis designed and it consists 01 insulating walls ofnegligible heat capacity with a hole in thebottom a drain plug fitted with a light rod toco~er the hole and 10 eledric immersionhebters each of 5 V0N The tank has beenshown inthe adjacent figure For testing water

is lfilled in the tank at 20degC and heaters areSW1itchedon After 15 hour of switching on theheaters water starts boiling at normalat+osPheric pressure The water is thenre~laced by an equal volume of oil whoseboiling point is 220degC at normal atmosphericprbssure and its specific gravity is 07 Theengineer reported that th~oil took 315 hour toraise its temperature from 20deg C to 60deg C at

normal atmospheric pressure

Immersion heaters

Drain plug

Figure Synthetic lubricating oil tank

From the report of the engineer the specific capacity of the oil is approximately found out(A) 60 of specific heat of water (B) 80 of specific heat of water(C) 90 of specific heat of water (D) equal to specific heat of waterA

1 is suggested that the air in the tank above the oil can be pressurized at 4 atm above normal airpressure which of the following is incorrect if pressure is increased to the suggested value(A) The time required to heat the oil would be greatly extended(B) The drain plug would be more difficult to lift(C) Fluid velocity would be increased when the tank is drained(D) The tilT)e required to drain the tank would decreaseAIn case of water (5x 10~ x 10 x 15x 3600 = V x PIN X Syx (100 - 20)In case of oil (5 x 1o~ x 10 x 315 x 3600 = V x Po x So x (60 - 20)After solving SrlS= 06 and V x Po = 57 x 103 kg

The amount of oil in the tank is nearly equal to(A) 12 x 103 kg(C) 36 x 103 kg-8

(8) 57 x 10~kg(D) 48 x 103 kg

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Paragraph for Question Nos 11 to 19

R B~0-1g

A space shuttle is designed in such a way that magnetic forceis used to reduce its acceleration during the landing of theshuttle on a newly discovered planet For this planet it wasknown that gravitational acceleration above the surface of theplanet is constant towards perpendicular downward to theplanet surface and is equal to g and the magnetic fieldproduced by the planet is uniformly B and is directed parallelto the planets surface There is no atmosphere above theplanet surface The space shuttle is given such a shape suchthat it can be idealized like conducting disc of mass m radiusR and thickness d (thickness d is much smaller than R) as nnnn7ne~aJlyjJY9J9)nnnnshown in the figure x and yare two flat surfaces of the shuttle Neglect any type of eddy currentappeared during the motion

17 The amount of charge induced on flat surface (say x) of the shuttle during the falls in thegravitational and magnetic field of the planet in terms of the shuttle speed v is(A) conR

2Bv (B) 2coR2Bv

(C) f-onRdBv (D) conR3B2v

The magnetic force acting on the shuttle during its fall in termSof the shuttle acceleration a is (A) conR

2Bda (B) conR2B2d ~)_

(C) 2conR2B2da (D) conR

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

The acceleration of the space shuttle during the fall is

9BconR2d1+-----

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SECTION - 8

Column-II(p) Work done by the system is

positive

(q) Work done by the system isnegative

II Increament in the internal(r) energy is equal to the heat

I transferred to the system

i----~ 7 -iI-~-- ~ 2~-d-l2 Y -~_-_-_- LILA -_-7 -~17v ~ ~-~ -o~~ --A--- - ---_q6--o-~-ampIffffffff

The conducting cylinder containing gas andenclosed by movable piston The cylinder issubmerged in an ice water mixture The piston isvery slowly pushed down from position-1 toposition-2p

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

i~ ltgt-- 0 - - ---~6-- --l-S-~rrrrFFFF ~

h The conducting cylinder coptaining gas andenclosed by a movable piston The cylinder issubmerged in an ice water mixture The piston isquickly pushed down from position-1 to position-2

Matrix - Match Typehi1s section contai~s 3 questions Each question contains ~atements givenIn two columns which have to be matched The statements 10 Column I arelabelled A B C and D while the statements in Column II are labelled p q r A

s ~nd t Any given statement in Column t can have correct matching with BONE OR MORE statement(s) in Column II The appropriate bubblesco~resPOnding to the answers to these questions have to be darkened as cillustrated in the following exampleIf t~e correct matches are A - p sand t B - q and r C - P and q and D - s Dand t then the corred darkening of bubbles will look like the following

1 Column-I contains some thermodynamic system and column-II contains its properties Match thefollowing columns

Column -I(A) Ametallic cube placed in open atmosphere and

gets heated from surrounding1

pOrT0I2 To T

A monoatomic gas follows the given process Theprocess 1 to 2 follows the thermodyn~mic relation

Tlt= Tg PPo

Increament in the internal(s) i energy is less than the heat

transferred to the system

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

transferred into the

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2 Column-I contains some charged system and column-II contains its properties Match thefollowing

Column -I Column - IIChargeSystem Properties

(A) z (p) Electric filled at the origin is zeroy

Electric field at z-axis Ez is alongpositive z-direction for zgtO and Ez isalong negative z-direction for z lt 0

+++++++

A ring placed in x-y plane symmetricallyhaving equal and opposite uniformly

distributed charge as shown in the figurey(B)

A s4uare placed in x-y planesymmeW~lIy ~aving a~ternate positive Iand negatIve umformly dlstnbuted charge ron each side of the square as shown inthe figure

-+~p p x

(r) Electric field atpositive x-direction

z-axis is along

Four identical dipoles placed symmetricallywith respect to origin in x-y plane as Ishown in the figure f

(0) z XX (s) Electric potential at any point on z-axis is zero

Electric potential at any point on z-axis is some nonzero positive value

Sol (A) -~ (r 5) (B) -)- (p s) (C) -)- (p q t) (D) -)- (r s)

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3 Column-I contains some circuits and column -II contains some properties In column-II currentsignifies instantaneous current and potential difference refers instantaneous potential difference

Column - I Column - Charge System Properties

x~E =amposin(rot)

A closely sepltllratedconducting wire iswrapped over an insulating cylinder Wireis connected to a resistance R and analternating source via a switch S which isclosed at t = 0

(p) In branch x current varies

(B) (q) Between points a and b potentialdifference decreases always

A charged metallic sphere of radius r isearthed with a resitance R and a switchS which is closed at t =0

In branch x current is constant

Between points a and b potentialdifference is constant

II (r)

r ~ St-NV( -

An inductor L and resistor R is connectedto a cell of emf c and internal resistance rwith a switch(s)which is closed at t = 0r~ ~I

I -~ s~1VwJ

a b A circuit (used to measure an unknownresistance) is in balanced conditions(Galvanometer reads zero) I

I (t)Rate of heat generation through theresistor R is constant

Sol (A) ---)(p) (B) --+ (p q) (C) ---)(P q) (D) ---)(r s t)

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Chemistry PART - II

SECTION-A

Straight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9 Each question has 4 choices (A) (B)(C) and (D) out of which ONLY ONE is correct

1 The number of alkenes (open chain as well as cyclic) of six carbon giving same product with HBrin presence of peroxide and absence of peroxide(A) 8 (B) 6~7 ~3

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

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C IDiethyl ether is very less water soluble but Tetrahydrofuran is completely soluble due to(A) Aromatic nature of tetrahydrofuran(B) Due to Sp3hybridisation of diethylether(C) Flipping of ethyl group prevents H-bond in diethylether(D) Diethylether stability in comparison of five membered ring intetrahydrofuranCFind out the correct match for the following compounds regarding their cis and trans configuration

(C) (D)

ir--I~CH3l~

(A) A - Cis B --trans F - Cis(C) C - Cis B - Cis 0 - CisBBecause

(F) f)jC~CH~I -

H(B) A - Cis B - Trans E - Cis(0) B - Trans E - Cis F - Cis

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A - Cis B - Trans C - CisD - Trans E Cis F - Trans

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4 Two drugs shown below areorally by the patient

F_--- CHO

I(( ---CH- C-OH- I

known as non-steroidal anti-inflammatory drugs (NSAIDS) taken

(Ansaid) (Mortein)(I) (II)

Which of the following is true(A) I is more soluble in H20 than II (8) II is more soluble in H20 than 1(C) Both are equally soluble in water (D) Both are insoluble in water

Sol ABecause of presence of F it develops H-Bonding

produced when propanoic acid is subjected to the following sequence of5

The amino acidreagents(i) PBr3 Br2(ii) H20(iii) NH3 1(A) Alutamic acid(C) Valineo

o degII PSr IIH~C-CH2-C-OH-~~) HbullC-CH-C~Sr- 6 - I

(B) Aspartic acid(D) Alanine

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

6 Find the correct statement(A) 2 4-pentanedione exists more as enol content in hexane than inwater(B) 24-pentanedione exists more as enol content in water than in hexane(C) 24-pentanedione is equally stable in water and hexane in keto form(D) 24-pentanediene exists mainly as keto form in water than in hexane

S I ADue to intramolecular H-bonding

7 Among the compounds

H)CH amp(i) (ii) NO

The correct order of pKa values are(A) IV lt III lt II lt I(e) IV lt II lt III lt I

(B) IV lt I lt II lt III(D) IV lt III lt I lt II

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r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

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8 In the reaction sequence

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

TOS)1ahor a I~S- hrdeted OH 00 bullbull

(ro~t~CD 0-8u--BuOH

cLgt9 The relative reactivities of compounds A Band C with HBr is

CH- CH- CH-I ) I ) I )H~C==C H-C==C HC==C

OCH3 CH3 CH2-O-CH

(Ai (8) (C)(A) A gt C gt B (B) C gt A gt B(C) B gt A gt C (0) A gt B gt C

Sol 0tn fact this is the order of stability of carbocation fanned after attack of etectrophile H-

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Reasoning Type

This section contains 4 questions numbered 10 to 13 Each question contains STATEMENT-l (Assertion)an~ STATEMENT-2 (Reason) Each question has 4 choices (A) (B) (C) and (D) out of which ONLYO~E is correct

o STATEMENT-l In the cytosineN-2 is better nucleophile than N-1 arxJN-3

H3C~I (111)~~STATEMENT-2 In the thymine lG~ amp cytosine N-2 of cytosine is better

N 0 nucleophile than N-2 of thymineH

(A) Statement-l is True Statement-2 is True Statement-2 is a correct explanation forStatement-l

(B) Statement-l is True Statement-2 is True Statement-2 is NOT a correct explanation forStatement-l

(C) Statement-l is True Statement -2 is False(D) Statement-l is False Statement-2 is TrueB

N 1-Statement 1 compare basicity of I in same molecule while in second statement the basicityis compared between two different moleculesIn cytosine lone pair of N-2 is not in conjugation while lone pair of N-l and N-3 in the same areparticipating in mesomerism In thymine lone pair of both N-l and N-2 are taking part inmesomerism

STATEMENT-1

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STATEMENT-2

STATEMENT -2(A) Statement-l is True Statement-2 is True Statement-2 is a correct explanation for

Statement-l(8) Statement-l is True Statement-2 is True Statement-2 is NOT a correct explanation for

Statement-l(C) Statement-l is True Statement -2 is False(D) SV~tement-l is False Statement-2 isTrueA

STATEMENT-l Cinnamaldehyde cannot go self aldol condensationandSTATEMENT-2 It does not have a-hydrogen(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

Statement-l(8) Statement-l is True Statement-2 is True Statement-2is NOT a correct explanation for

Statement 1(C) Statement-l is True Statement -2 is False(D) Statement-1 is False Statement-2 is TrueCIn Cinnamaldehyde (PhCH = CH - CHO) there is no acidic a-hydrogen

eg PhCH = CH - CHO~PhCH = C~ CHO

The carbanion is not sufficiently stabilized

13 STATEMENT-l When mixture of 1 mole of benzaldehyde and 1 mole of cyclohexanone istreated with semicarbazide semicarbazone of cyclohexanone precipitates first and finallyprecipitate of semicarbazone of benzaldehyde is formed I

andSTATEMENT-2 Initial one is kinetically controlled product and the other one isthermodynamically controlled product(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

Statement-l(8) Statement-l is True Statement-2 is True Statemenl-2 is NOT a correct explanation for

Statement-l(C) Statement-l is True Statement -2 is False(D) Statement-1 is False Statement-2 is True

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Comprehension Type

II~ITS-PT -110 Paper -1 -PCM(S)-JEE(AdvancedV13 m roxaxz

I Tlhis section contains 2 groups of questions Each group has 3 multiple choice question based on aP1aragraph Each question has 4 choices (A) (B) (C) and (D) for its answer out of which ONLY ONE iscprrect

1 OK (excess)~ ) G-CH~2 K

(yellow precipitate)

1 (C03J 00 ekOIo~ bullC

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

Compound j iso 0II 1

H-C-C-CH -CH-C-O-CHgt 2 I 3

Compound B is(A) It is pleasant smelling red coloured liquid(B) Sold in paint stores for general clean up work(C) It is less volatile than water(D) All the above

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

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CH~o 0~ l

(ElHoC-C--CH-C-O-CH _ 3

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(G) HO-C-CHoI bullCH3oII

(I)HC-O-C-CH3 (J)

o 0II II

H C-CH -C-O-CH-C-O-CH3 2 I 3

o adation A[C H ] H3eq) [ ]CH3CO H+ E (KM~O 9 12 Pd-C) B C9H~6

jaq H2S04

C+D(isomeric ketohes)

O H2C-C-OHO

II I IIIf E__ CH C CO andcompoundC gives positive iodoform testT-CH2- 2- H2- 7- H

OHCompound B is(A) CH3

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

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Solution for the Q No 17 to 19

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

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SECTION-B(Matrix Type)

This section contains 3 questions Each question contains statements givenin two columns which have to be matched The statements in Column I arelabelled A B C and D while the statements in Column II are labelled p q rsand t Any given statement in Column I can have correct matching withONE OR MORE statement(s) in Column II The appropriate bubblescorresponding to the answers to these questions have to be darkened asillustrated in the following example

If the correct matches are A - p sand t B - q and r C- p and q andD - sand t then the correct darkening of bubbles will look like the following

1 For each of the statements in Column I choose a substituent from Column II that fits the

descriptionfor thecompound~z

ColumnJ(A) Z-donates electrons inductively but does not donate or

withdraw electrons by resonance(B) Z-withdraws electrons inductively and withdraws

electrons by resonance(C) Z-deactivates the ring towards electiophile and directs

ortho para positions(D) Z-withdraws electrons inductively but donates electrons

by resonance and activates the ring

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

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I Cope elimination (a typical syn-elimination)General mechanism

R1041+

$~~3 Match the following

Column-I (Monocyclic species where Iall carbon atoms are part of cycle)

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

Column-ll (Properties)

Contains 2 n-electrons

Contains 8n-electronsContains 10n-electronsAromatic in nature

Sol A ~ s t B ~ r C ~ p t D~ q t

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

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PART - III

SECTION -A

This section contains 9 multiple choice questions numbered 1 to 9 Each question has 4 choices (A) (B)(C) and (0) out of which ONLY ONE is correct

px + qy = 40 is a chord of minimuml~n~th of the circle (x - 10)2 + (y - 20i = 729 If the chordpasses through (5 15) Then (p2013+ qo) is equal to(A) 0 (B) 2(C) 22013 (0)22014

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

503 Only one common tangent

3 The number of ordered pairs (x y) satisfying Ixl+ IYI = 3 and sin( 2x2 = 1 isare~ 3 j

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

f3 15=gt x = t~2 t~-2- and for each x there are two values of y

The number of ordered pairs = 8

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4 If the asymptotes of the hyperbola (x + y + 1 - (x-y - 3)2 = 5 intersect at A and the coordinateaxis at Band C then the equation of the circle passing through A Band C is(A) 4(x2 +y2)~4x+6y+3 == 0

(C) x2 + l -x + 2y == 0Sol C

Centre of rectangular hyperbola is (1 -2) Equation of Asymptotes are x = 1 Y =-2

Radius of circle = J5 2

(6) 4(x2 +y2)-4X+y 0

(D) none of these

Let PQRS be a square with length of the side 10 The points A and 6 are taken on the side PQand C is taKen on the side RS Then the circumradius of the triangle ABC can be(A) 31 (6) 42

gt (C) 55 (D) 8CLet M be the circumcentreNow MA + MC ~ AC ~ SP = 10 The minimumvalue of circumradius = 5Now a circle through PQRS has the radius 5J2 and the points P- Q R lie on or inside the-square

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

In triangle ABC with A as obtuse angl~_AD and AE are median and~ltitude respectively If

~AD == IDAE == IEAC then sin3 A cos A equals 3 3(

(A) 3a3

(C)3a3

256b2c6Clearly triangle ADC is isosceles= AD = AC = b

=gt DE = CE ~ ~4

Now sin3 A cos ~ == sin2 A 2sin A cos A3 3 2 3 3 3

1 2 A 2A= -Sin -sln-2 3 3

1 (EC)2 (BE 1 a2 3a 1 3a3= 2 b IAB == 216b2 4c == 128b2c

(A) (-11)

(C) RCx2 + y2 1 lYIJ2 and Ix - 21~ 1No such (x y) exists K E R

If sin (x2 + y2) + tan Jy2 - 2 + sec(x - 2)= K has no solution then exhaustive range of K is

( 1 1 (6) 1--== -1

--2 J2)(D) none of these

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This section contains 4 questions numbered 10 to 13 Each question contains STATEMENT-1 (Assertion)and STATEMENT-2 (Reason) Each question has 4 choices (A) (B) (C) and (D) out of which ONLYONE is correct

D Y C Y

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ab - rr STATEMENT 1 In tnangle ABC 12 == R

STATEMENT 2 In any triangle ABC _a_ = _b_ =_c_ = 2R sinA sinB sinC

(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(D) Statement 1 is false and Statement 2 is trueoab = 4R2 sin A sin B

rh = 4R2 sinAsinBcos2 ~2

The area enclosed by the curve max Ix - 11WI = kis 100 then Iltis equal toWS ~8(C) 10 (D) none of theseAArea enclosed by the regionABCD = 2k 2k=gt 4k~ = 100k=S

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

Equation of tangent PT is ty = x + atEquation of normal at a is xt2 - yt3 - a - 2ae = 0 PT II aNd = l-at4 - a - 2ae 1= ah

2 + 1)32

)t4 + t6 t2

P is a point on the parabola y2 = 4ax and PO is a focal chord PT is a tangent at P and ON is anormal at O The minimum distance between PT and ON is equal to

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

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STATEMENT 1 The normals at the ends A and Bof a focal chord of the parabola y2 = 4ax meetthe parabola again at C and D respectively then AB = 3CD because

STATEMENT 2 The normal at t meets the parabola again at t2 then 12 = -t - 3t

(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false (D) Statement 1 is false and Statement 2 is trueA

2 (a 2aLet A = (at 2at) B = 12 - -t )

Also normal at A meets the parabola at C 2 2 ) (2 2 (1 1

c=lai-t--) 2a(-t--lJandD=la-+2tl2al-+2tl t L ~t t

bulllI

10Ts-P1 -HPaper-1 l-PCM(S)-JEE(AdvancedV13 27

1- ST~TEMEN~1 If xoxl are the 19th of the segments of the focar~~~IiPse whoselatus rectum is given by 2(x) Then major axis of the ellipse cannot exceed 2STATEMENT 2 Semi latus-rectum is the HM between the segments of the focal chord(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(D) Statement 1 is false and Statement 2 is trueD

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

1 STATEMENT 1 In any triangle ABC the minimum value of cot2 A + cae B + cotL C is J3 STATEMENT 2 y = cot x is a concave downward curve in 10 n)(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(D) Statement 1 is false and Statement 2 is true

I Comprehension Type

This section contains 2 groups of questions Each group has 3 multiple choice question based on ap~ragraph Each question has 4 choices (A) (B) (C) and (D) for its answer out of which ONLY ONE is

I ctgtrrect

(B) Xi + 4 = 9(D) none of these

Read the following write up carefully and answer the following questionsBly taking B(-3 0) and C(3 O)triangles PBC of perimeter 16 are constructed If S is the locus of incentreof triangle ABC then

Equalion o S is(A) xlt + 4yl = 4

I C)x2+4l=12SOl B

As ID -LBC

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17 The equation of S is

Paragraph for Question Nos 17to 19

Read the following write up carefully and answer the following questionslet 0(0 t) is a variable point andR = (3 0) PO is drawn perpendicular to RO to meet the variable line y =2t in P let S represent the locus of P

AITS-PT -1I-Paper-1)-PC M(S)-JEE(Advanced)13x nY ytm-

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

Twocperpendicular tangents to S inter-sect at O then 1001 is equaf to ( being origin)~

(A) ~2c

(C)~-j32

A ray emanating from (- 33 0) is incident on the curve S at P whose ordinate is The

equation of the reflected ray is(A) 4x + 313y = 12

(C) 8x + 313y = 1213CEquation of s is x2 + 4y2 = 9

p (0~I2

Foci are 1- -0 I2 )Equation of the reflected ray is 8x + 3J3y = 1213

Now x = BD - BO = 5 - b ~ b = 5- xC=10-b=5+x----------Now j = ~8(8 - 6)(8 - 5 + x)(8 - 5 - x)

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

Equation of PO is Y - t = ~(x - 0)t

Coordinates of P = (~ 2t]3

locus of P is y2 = 12x18 The chord x + y = 1 cuts the curve S in points A and B The normals at A and B intersect at C A

third line from C cuts the curve S normally at O Then the coordinates of Dare(A) (3 ~) (B) (0 0)(C) (6 -3) (D) (12 12)

UWttrfPY X HlOO )

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How many nonnals can be drawn from the point (J37 -10) to s(A) 1 (8) 2(C) 3 (0) infiniteCAs h gt 2a 3 nonnals can be drawntos

Let A (3m~ - 6~) B (3m~ - 6m2) 0 (3m~ - 61113)A and B lies on x + y = 1 3m~ -6m1 =0 and 3~ -6m2 =1

m - m~ - 2( Ill - m2) = 0Ill + m2 = 2 ~ --m3 = 2 Coordinates of 0 are (12 12)

) AIS-PTjPr~CM(S)-JEE(Advanced)13

29taN we xh Yilt- OX xrtYrr raquonW) r( tTlC- (

SECTION-B

(Matrix Type)

Ttlis section contains 3 questions Each question contains statements givenin Itwo columns Which have to bematched The statements in Column I arela)elled A 8 C and 0 while the statements in Column II are labelled p q rs ~nd 1 Any given statgmentin Column I can have cOJrect matching withONE OR MORE statement(s) in Column II The appropriate bubblescoirresPOnding to th~ aQ~wers to these questions have 19 be darkened asillustrated in the foHowing example

I If the correct matches are A - p sandt B - q and r C - P andq and 0 - s~jd tthen the correct darkening of bubbles will look like the following

p q r s

ACB900E9B 0 ~EfD0c)c Ci)~~)0C0D~00CDCD

(A) ~ (r) (Bl-~ (r) (C) ~ (p) (D)~ (q)The 4 points of intersection lie on 2x2 + 12x - By + 26 = 0 (by adding)Which is a parabola with vertex = (-3 1) S = (-3 2) and directrix y = 0(A) A = 0 8 = 0 g = 3 f = -2 c = 13 ~ g + f + c = 14(B) a=-3p=2 a2+132+1=14(C) Required sum = sum of these points from y = 0 = sum of the ordinates of these points

These points of intersection also lie on l- 20y + 59 = 0 Sum of ordinates = 20 (By subtracting)

Since these points lie on a circle and hyperbola so we get a pair of points with same ordinatesHence sum = 40 -(0) As SG = SP = 5

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2 Match the following Column-I with Column-II

Let P be any point on the parabola = 8x between its vertex and (p)positive end of latus rectum If 5 be the focus and Nbe the foot ofperpendicular from 5 to tangent at P then possible integral valuesof area of triangle PN5 isA chord is drawn from a point A(1 a) to the parabola l = 4x which (q)cuts the parabola at P and Q such that APAQ = 31al then ex cantake the valuesA trapezium is inscribed in the parabola l = 4x such that its (r)

diagonals pass through the point (1 0) and each has length 25 If4

the area of trapezium is A then 16 A is equal to25

Normal to the parabola yl = 4x at points P and Q of the parabola (s)meet at R(3 0) and tangents at P and Q meet at T (a 0) thenarea of the quadrilateral PTQR is equal to

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

(B) Equation of the line thro (1 a) is x -1 = Y- a = r It meets the parabola at P and QcosO SIOO

~ (rsinO +o = 4r(rcos6 + 1) ~ r2 sin2 H+ 2r(tsinO t2cose) + (a2 - 4) = 0

~ Ap x AQ == 1a2

-2 41~3lalsin e-

la2 -- 41 2~ 31al - sin es 1 ~ (Ial + 1)(101- 4) ~ 0 ~ lal ~ 4

(C) The diag0lals arc the focal cords of the parabola RPF = 1 + tlt = a (say)

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

For R t = t1 xATt=1a=-1 (aO) PM = 2at = 2~ RT= 4

Area of quadrilateral = i x (i-4 x 2) = 8 sq units

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If the cirde x2 + y2 - 4X - 6y + k = 0 does not touch orintersect the axes and the point (2 2) lies inside thecirde then k can lie in the intervallintervalsIf the line 3x + ay - 20 = 0 cuts the circle x2 + y2 = 25at real distinct or coincident points then a belongs toIf the point (2 k) lies outside the circlesx2 + y2 + X - 2y - 14 = 0 and x2 + y2 = 13 then kbelongs to the intervalIf the point (k + 1k) lies inside the region bounded by

the curve x = ~25 - y2 thenk belongs tothe interval

(p) (-13)

(q) (-Y -3)v( 4x)

(r) (~J-j7Jv[ -fioo)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

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2 0 C B A

3 ABC A

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5 A ADD

6 A A 0 C

7 B A A C

ANSWERS HINTS amp SDLUTIONSPART TEST II

PAPER2

Physics

oABo(A) ~ (5)(B) ~ (r)(C) ~ (q)(0) ~ (p)(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(A) ~ (5)(B) ~ (p)

I (C) ~ (q)i (0) ~ (r)

o(A) ~ (5)(B)~ (p)(C) ~ (q)(0) -4 (r)(A) ~ (5)(B) ~ (r)(C) ~ (q)(0) ~ (p)(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)

PART -I

oooABAA

o(A) ~ (5)(B) ~ (r)(C) ~ (q)(0) ~ (p)(A) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

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o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

AilS-PI ~II-Paper-2)-PCM(S -JEE(Advanced132

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

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10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (p r) I (A) - (p r)(8) ~ (p q) I (8) - (p q)(C) ~ (q r s) I (C) - (q r s)(D) - (p q s) (D) - (p q s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

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-PART -

SECTION-A

1 bull

~YSICS

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Th1issection contains 9 multiple choice questions Each question has four choices (A) (B) (C) and (D)oul of which ONLY ONE is correct

2 In the figure Box-1 contains one capacitor and one resistorBox-2 contains 1 inductor and one resistorR = resistance in Box 1R2 = resistance in Box 2

1Power factor of Box 1 = -r

Power factor of Box 2 = - r=05

1Power factor of Box 1 + Power factor of Box 2 = ----=

Vosin(rot)

Power factor of Box 3 = -----=J2

Resistance of Box 3 = R + R2

Current I = a sin(wt + ~)On the basis of above mentioned information pick the incorrect option from below

(A) ~ = 3 (B) Z3 = 4i2R2 (Impedance of box 3)Rz(C) 1= Vo sin(Cilt) (D) (j = (2n + 1)n where n may be some integer

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Box 1 1--ffiC1 1roC-=2R

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

1-- - rolffiC1 = 1R1 +Rz

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

Impedance of Box 1 + Box 2 = (R + Rz)12 = 4--J2 R2 phase angle is 1t4

Impedance of Box 3 = 4--J2 R2 phase angle is -n4

Current in Box 1 + Bo~ 2 = I = J sin(wt ~ ~)42Rz 4

Vo ( 1t 1Current In Box 3 = 12 = r SIAl wt - -

42R2 4

= I + 2 = Vo sin( (J) t)4R2

I d f B 1 B 2 B 3 Vosinffit 4R d tmpe anceo ox + ox + ox = ---= 2 an IS resistive In na ure

Phase angle is zero

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6 AITS-PT-U~Pape~-2)-PCM(S)-JEE(Advanced)13it O

-11+41-3J3ilt E = J44

In the figure OA and OB are two uniformly chargedwires placed along x and y axis respectively The unitvector in the direction of field at point (0 0 -Y3L) is

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

4 In the figure A B C and D are four concentricspheres of radius a 2a 3a and 4a and containcharges +3q +6q -9q and +12q respectively In thebeginning switches 81 and 82 are open After earthingthe sphere C Ilq1 amount of charge flows from thesphere C to the earth If the sphere B is also earthedthen Ilq2 amount of charge flows from the sphere B to

the earth the value of IIlq1 I is equal toItiq2 I

(A) 1(B) 35(C) 23(D) 12

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_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

1 [3qbull q2 q3 12q1 041tco 3a + 3a + 3a + 4a J =_1_ [3q + q2 + ~ + 12q] = 04mo bull2a 2a 3a 4aSolvingq2 = -3ql~q21= 9qI~q2 I = 1ILq I

(8) 5 ms(D) 2 ms

v = -vx + C~ v = vo -- Vx = vo - v sin q

~ v = --~ = ~ = 3ms1+sin~ 1+1

A charged particle having charge q and mass m is projected into a region of uniform electric fieldof strength Eo with velocityVo perpendicular to Eo Throughout the motion apart from electricforce particle also experiences a dissipative force of constant magnitude q Eoand directed

opposite to its velocity If I Va 1= 6 mis then its speed when it has turned through an angle of 90degis equal to(A) 3 ms(C) 4 msAConsider the situation when particle hasturned through an angle of ltjJ

dv = _ qEo [1- sin$]dt mdvx _qEo[1 ]---- -Sindt m~ dv = _ dvx

dt dt~

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2 2B2(B) ~-q-

6 A uniformly conducting wire is bent to form a ring of mass m and radius r and the ring is placedon a rough horizontal surface with its plane horizontal There exists a uniform and constanthorizontal magnetic field of induction B Now a charge q is passed through the ring in a very smalltime interval Llt As a result the ring ultimately just becomes vertical Calculate the value of g(acceleration due to gravity) in terms of other given quantities Assume that friction is sufficient toprevent slipping and ignore any loss in energy

(A) ~ (r r

2 (q2B2r)(C) 1t --

Angular momentum (~mr2)()= r ldt = f dq nr2Bdt == q1tr2B2 dt

For the ring to become vertical1(3 212- -mr ill = mgr2 2 j2( B2 1t qIe g=- --I r

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

where r = 2R cos 4gtv = r_1~dq 4mo r

A nQn-conducting wire is bent into a semi-circle of radius R and achafge + Q is uniformly distributed o~er it as shown in the figure

Find ithe ratio (~ J of the potential (VA) to the magnitude of the

electric field (E~ both evaluated at the other end A of diameter ABthat bisects the wire (as shown in the figure)(A) R(C) R In (1+12)

B 1 dq

E = I---cosoo 4nso r2

8 In the figure ABCDEFA is a closed conducting wire loop2 ~

The part AFE is a portion of an ellipse X 2 + y = 1and4a a

BCD is a portion of the ellipsex2 y2 X-+-=19a2 4a2

A uniform magnetic fieldB == Bok exists in this region If the frame is rotated aboutthe y-axis with a constant angular velocity r) Themaximum emf induced in the loop is

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1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

Ca = 2aa = 3aArea of wire frame

n(a2bpound - alb) 2-----=21ta

2ltl = 21tBoa2cos(lJ)t

C = - dljl = 2ll(J)Boa2sin (rot)

peak emf = 211lJ)80a2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

In the figure 8 B2 and 83 represent uniformtime vaJ)ing magnetic fields in three differentinfinitely long cylindrical regions with axispassing through P Q and Randperpendicular to the plane of the paperPQR is an equilateral triangle of side 43 rwhere r is the radius of each cylindricalregion (r = 1m) The magnetic inductionsvaJ) with time t as B = A+ at 82 = A - btand B3 = A + ct Here a = 2 Testas b = 2Teslals and c = 1 Teslas and A is aconstant with dimensions of magnetic fieldand t is time The induced electric fieldstrength at the centroid of the triangle PQRis

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

CThe contribution due to anyone thecylinder

IE 1=~ldBI = CldBI = ldBI2x dt 8r dt 8 dt

IEI =V 1m4

IEzl = 14 VImIE31 = 18 VIm(directions as shown in the figure)1- - - 13IE + E2 + E31 =8 VIm

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

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10 AITS-PT -11-( Paper -2)-PC M(S )JEE(Advanced)13nrx

Assertion - Reasoning Type

STATEMENT-1A constant potential difference is maintained between the inner andouter surfaces of an annular homogenous and isotropic conductingsphere of inner radius a and outer radiusb The drift speed ofcharge carriers (Le electrons) decreases with increase in distancefrom the centreand

This section contains 4 questions numbered 10 to 13 Each question contains STATEMENT-1(A~sertiOn) and STATEMENT-2 (Reason) Each question has 4 choices (A) (B) (C) and (0) out of

Iwhlich ONLY ONE is correct

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mCompressedmonoatomicideal gas 5mole 400 K

STATEMENT-1The adjacent figures shows two thermallyinsulated vessels having equal volumes andcontaining monoatomic and diatomic ideal gaSeurosseparated by a stopper which prevents thegases from mixing with each other If the stopperis removed and the gases are allowed to mixthe final equilibrium temperature is 525 K andSTATEMENT-2Total energy of the system is conserved

STATEMENT-1 In the circuit shoWh in the figure initially S~ is closed and S is-open After some time S2 is closed and S1 is openedsimultaneously The ammeter reading shows a sudden change

andSTATEMENT-2By Faradays Law

em = - dltp A- = magnetic flux = Induced emf dt t

(C) Statement-1 is True Statement -2 is False(0) Statement-1 is False Statement-2 is TrueB

STATEMENT-2As we move away from the centre the electric field decreases in magnitude (A) Statement-1 is True Statement-2is True Statement-2 is a correct explanation for

Statement-1(B)Statement-1 is True Statement-2 is True Statement-2is NOT a correct explanation for

Statement-1(C) Statement-1 is True Statement -2 is False(0) Statement-1 is False Statement-2is TrueA

I501112

50[111

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(A) Statement- is True Statement-2 is True Statement-2 is a correct explanation forbullStatement-1(B) Statement-1 is True Statement-2 is True Statement-2 is NOT a correct explanation for

Statement-(C) Statement- is True Statement -2 is False(D) Statement-1 is False Statement~2 is TrueA

13 STATEMENT-1A metallic sphere A has radius R and a concentric hollowmetallic sphere B has radius 3R The space between them isfilled by an annular dielectric of dielectric constant 2K andradius R (inner) and 2R (outer) and another dielectric of dielectric constant K and radius 2R (inner) and 3R (outer) Nocharge is deposited at the common interface of the dielectrics and

STATEMENT-2

Asper Gausss law ~EdS = ~E(Symbols have their usual

meanings(A) Statement- is True Statement-2 is True Statement-2 is a correct explanation for

Statement-1(B) Stt~ment-1 is True Statement-2is True Statement-2 is NOT a Gorrect explanation for Stafement-1 r i~

(C) Statement-1 is True Statement -2 is False(D) Statement-1 is False Statement-2 is1rue

Comprehension Type

This section contains 2 groups of questions Each group has 3 multiple choice question based on aparagraph Each question has 4 choices (A) (B) (C) and (D) for its answer out of which ONLY ONE iscorrect

Four large metallic plates of area A each are kept parallel toeach other with a small separation between them as shown inthe figure A cell of emf V is connected across the twooutermost plates through a switch K1 The two inner plates aresimilarly connected with a cell of emf 2V through a switch K2bull

Initially both switches are open and the plates starting from leftto right (ie number 1 to 4) are given charges Q 2Q -2Q -Q respectively Now answer the following questions

1 2 3 4

14 The charge appearing on the outer surface of plate -1 when switches K1 K2 are open(A) zero (B) Q(C)-Q (D)-3Q

Sol AWhen both switches are openCharge on outermost surface = (net charge on system)2

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12 AITS-PT -IIi Paper -2)-PCM(S)-JEE (Advanced)13

I15 If K1 is closed and K2 is open the charge appe~ring on the right surface of plate-2 is

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

d 4 2BWhen K1 is closed and K2 is open

9 + ~ + ~ = V where Co = eoACo Co Co12 dFor plate 2q - q1 = 20Solving the above equations we get

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

If both switches are closed the charge appearing on the plate 4 is

(A) (C~A ) V (B) 0 + C~A (~) 2t7

(C) 0 - COA( Vl (D) (EoA Y Vjd 2) d ) 3 j

DWhen both switches are closed V

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

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A calorimeter of water equivalent 250g contains a block of iceat -1 Dec in which there is a solid conducting ball whosewater equivalent is 50g The mass of ice is 800 g 150g ofsteam at 1000e is added to the calorimeter and the system isisolated from surroundings calorimeter

xtoxrmraquoraquox

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AIlS-PT -Iii Paper -2)-PCM(S)-JEE(Advanced)13ret ~ laquoftt0

GivenSpecific heat of water = 1 calgoCSpecific heat of ice = 05 calgOCLatent heat of fusion of ice = 80 callgLatent heat of vaporization = 540 calg Answer the following questions based on the infonnation above

17 Amount of water in the container when equilibrium has been established is(A) 950g(Bl800 9(C) 150 9(0) cannot be calculated from the information given

Equilibrium temperature of the system is(A) Dec(C) 20CCC

(B) 10QoC(0) 30iit

19 The density of the material of the ball at OC is Ip6(OdegC)is equal to density of fater Also PB

varies with temperature as PB(Gec) = PB (OdegC) x r1-~) Ignore the expansion of water and any 200

pressure changes with temperature After thermal equilibrium has been reached the fraction ofvolume of the ball in air in the calorimeter equals

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

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(B) Six hundred mole of a gas undergoesa process (q) 27

P = 7 x 105 N 012

iii 2 +Y_4m3

where V is volume in 013 When volume of thegas is 20 01 its temperature in Kelvin isapproximately equal to

1 Column-I contains some thermodynamic system andjX)lumn-1I contains its properties Match thefollowing columns

Column - I Column - II(A) A steel rod is clamped at its two ends and rests (p) 400

on a fixed horizontal base The longitudinalstrain developed in the rod when temperaturerises to 50degC is - 36 x 104 and the temperaturecoefficient of linear expansion of steel is 12 x105 fc Find the temperature at which rod is inrelaxed state

100raquo0 O--=gtraquoW= 14 AITS-PT -t1-Paper-2)-PCM(S )-JEEl~dvanced)13y

I SECTION ~B

I Matrix - Match TypeT~is section contains 3 questions Each question contains statementsgi~en in two columns which have to be matched Statements (A B CD) in column I have to b~ matched with statements (p q r s) incolumn II The answers to these questions have to be appropriatelybtbbled as illustrated in the following example

If ~he com~ct match are A-p A-s B-q B-r Cop C-q and O-s then thecorrectly bubbled 4 x 4 matrix should be as followsl

(C) A block of mass 100 gm slides on a rough (r) 20horizontal surface of zero heat capacity Whenthe speed of the block decreases from 10 ms to5 mIs its temperature rises by O2SoC The heatcapacity of the block in joulecC is

(0) P(Nm1) (s) 15

5 3 x10

isoczj0riCprocess isothermalprocess

A- t isobaric B

process

T 600 T (Kelvin)If two moles of a gas consisting of 1 mole of Heand 1 mole of H2 undergo a cyclic processA ~ B~ C ~ A as shown in thefigure the totalwork done in the cycle is -600Rln[9e] Thetemperature T in DC is

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

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~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

Match the following columns- Column-I

Column-II(Temperature Tstate)sooe

in steady

Three homogenous thermal resistors A Band eeach of length I cross-sectional area A and thennalconductivity 2Ko 3Ko and 6Ko respectively are joinedin series such thatthere is one dimensional heat flowfrom left end(temperature 100degC) to the right end(temperature 1Q0C) T is the temperature of theinterface between resistor Band e in the steadystate

A spherical annular thennal resistor of thennalconductivity K and inner and outer radii 2R and SRrespectively is in steady state with its inner surfacemaintained at 100degC and its outer surface maintainedat 10degC (heat flow is radially outwards) Thetemperature of the concentric spherical interface ofradius 3R in steady state is T

ltSH--

bull 2R ~ T bull

A cylindrical and thennal resistor of inner radius Rand outer radius 8R is in steady state with its innersurface maintained at 100degC and its outer surfacemaintained at 10degC(heat flow is radially outwards)The temperature of the co-axial cylindrical interfaceof radius 2R in steady state is T

(r) IssoeI

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100degC 10degC

TA homogeneous thermal resistor whose outersurface is the frustom of a solid cone has the twobase radii as 6R and 3R respectively and theperpendicular distance between its two bases is 3LThe two bases are maintained at 100cC and 10degC Inthe steady state(heat flows from left end to the right end) the temperature of the interface at a distance of2L from the left end is T Assume that L is very largein comparison to Rand the lateral surface of thecylinder is thermally insulated from the surroundingsFindT

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

Column-I contains some physical systems and column-II contains its properties Match thefollowing columns

Column - I Column - II ~(A) A metallic solid sphere of mass 5 kg and radius 2 (p) 60

m is freely rotating about its diameter withangular velocity 00 = 2 rads The coefficient ofthermal expansion of the sphere isa = 104 fcIf temperature increases by 20degC the changein angular velocity of the sphere is

(B) A metallic sphere of emissivity 08 radius 4 m (q) I Zeroand at a surface temperature of 600 kelvin isemitting radiation If half of the metallic sphere iscovered by some opaque material and thesurface temperature increases to [600 (52) 14kelvin the change in power emitted is

(C) A metallic hollow spherical shell is placed (r) 25concentrically with a spherical shell of double theradius The inner sphere is given a charge qwhile the outer sphere is given a charge 4 qNow the two sphere are connected together by ametallic wire The change in charge on theouter surface of the outer sphere is

(D) Dielectricof (s) 04

q~ ~2 ~ ~t~c=c~n$tant

C ~ ~~ V q2 V

Situation-1 Situation-2Two identical air capacitors 1 and 2 ofcapacitance C each are connected in series anda potential difference V is applied across the

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AITS-PT -1I-(Paper-2-PCM(S-JEE(Advanced)13 17Ylt OX iXS(

combination as shown in the figure q1 is thecharge of capacitor-1 in situation-1 When adielectric of dielectric constant 4 is inserted intocapacitor-2 a charge q2 flows through the batteryin the circuit as shown in the figure in situation-2

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

ampeV2 eV2 4CVl5 4CV5~J~JYv Yv qz

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PART - n

AITS-PT-IHPaper-2)-PCM(S)-JEE(Advanced)13xooa raquoxi rrt tiC

5ECTION-A

Straight Objective TypeI

T~iS section contains 9 multiple choice questions numbered 1 to 9 Each question has 4 choices (A) (8)I

(q and (D) out of which ONLY ONE is correct _

(~ 11~

~r il~

How many total stereoisomers do the following compounds have H 00 CO 0(~) (B) (C) (0)

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

(E) None(F) Enantiomeric pair

~ __ ~~i

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2 How many products will formed by reaction

()0pounddeg4ii)piHSO~

(A) 4 (8) 3(C) 2 (D) 1

Sol CTwo -OH groups above and two below the plane

3 Which of the following will not exhibit geometrical isomerism

ltgt=lt1Cl

(D)(C) COoIn A and 8 there is hindered rotation about central bond between two rings

(A) I) lt I M L gt-ltJ (6)M 00

In (C) compound exist in two form cis and trans(C) H H

CbOOCbH H

In case of (D) on rotation conformer will develop instead of geometrical isomer

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

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In the followin table the value of di ole moments (in Deb e are as followsX CHX CH2X2 CHX3__ ~F a a2 a3 iCI _b___ ~ C3 I--------jBr __ c__ C2 C3

I d d2 d3The correct relation among a a2 a3 b ~~ Ch C2C3and d d2bull d3 are(A) a3gt a2gt a b gt ~ gt be (B) a2gt agt a3 b gt ~ gt b3(C) Cgt C2gt C3d2gt d gt d3 (D) C gt C2gt C3b2gt ~ gt bB

20 AITS-PT -1I-(Paper-2)-PC M(S )-JE E(Advanced)13urn lfrTMraquo)

consists of three fused chair conformations If 1167gms of the abovecompound undergoes acid catalysedhydrolysis it requires how manygrams of water for complete hydrolysis(At Masses C - 12 N -14 H - 1)

X CHX CH2X2 CHX3F 182 197 165 CI 194 160 103Br 179 145 102I 164 11 100

C~WM rtiNbull-_ bull--------N

(A) 5 (C) 15BgtIt is urotropin and its reaction with water is

(CH2)EN +6H20~4NHo + 6HCHO

(B)9(D) 10

0gt--- OHo ---lt)--J0H(D)

(C) HO~

7 In the cyclization reaction given below the most probable product of the reactionSr

~ A OH ~ bullProduct isHO ~ ~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

IOH group nearer to Br is more acidic

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8_ In the reaction amp(2 ampq)h

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

1 EtONalEIOH l Q~so (dOl) (Compound with six mombampampO nng)

P+Q 00-(00005) lFinalproduct-

NCliO-----I il-J~CH--Br

deg~OEtyOEtdeg(2 mole)

Then final product is

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

o~lH)l 8EtO + r OEt Et~ Na-

~OEI EtOHEt H~ V II-

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22 AllS-PT -1I-(Paper-2)-PCM(S-JEEAdvanced)13

OW (excessbullbull J

bullbull

o(This compqpnd may further ) undergo tautomerism to give

~ CHJCOOH + CO

(B) less volatile than water(D) non-aromatic compound

2HOo(6)

(8irc duct) )

A Na) B ~HO03) 2C~ 2CH3COOH 2C02[HydJOcarbon J liquid NH3 -

Then compound A is(A) colourless liquid(C) orange red gasAg

Compound A is benzene which is colourless liquid

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Reasoning Type

cordgt Wetc

10 STATEMENT -1 Acetaldehyde reacts with alkaline solution of sodium nitroprusside to give redcolourationandSTATEMENT-2 Acetaldehyde is good reducing agent(A) Statement-1 is Tru~ Statement-2 is True Statement-2 is a correct explanation for

Statement-1(B) Slatement-1 is True Statement-2 is True Statement-2 is NOT a correct explanation for

Statement-1(C) Statement-1 is True Statement -2 is False(D Statement-1 is False Statement-2 is True

11 STATEMENT -1 Ortho iodobenzoic acid is strongest acid among ali ortho halobenzoic acidandSTATEMENT-2 Iodine exerts maximum ortho effect so the acidic strength increases(A) Statement-1 is true Statement-2 is True Stafement-2 is a correct explanation for

Statement-1(B) Statement-1 is TruegtStatement-2 is True Statement-2is NOT a correct explanation fof

Ph-S CH2

IOHS is a better nucleophile than 0 atom and act as neighbouring group participant

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STATEMENT -1 Generally esterification reaction of carboxylic acid and alcohol is carried out inacidic medium while esterifIcation reaction of acid chloride and ~Icohol is carried out in basic-mediumandSTATEMENT-2 In acid chlorides and alcohols reaction HC is released which can be neutralizedby base(A) Statement-1 is True Statement-2 is True Statement-2 is a correct explanation for

Statement-1(B) Statement-1 is True Statement-2is True Statement-2is NOT a correct explanation for

Statement-1(C) Statement-1 is True Statement -2 is False(D) Statem-ent-1 is False Statement-2 is TrueA

- OXn- bullbull lUX 0 bullbull1lt9( ro- )yrtSO(((((UOUzy-xrxraquoXftYngtxp

24 AllS-PT -II-(Paper -2)-PC M(S )-JE E(Advanced)13Y M

Comprehension Type

This section contains 2 groups of questions Each group has 3 multiple choice question based on ap1ragraph Each question has 4 choices (A) (B) (C) and (D) for its answer out of which ONLY ONE is

c~rrect

llihen wate~ is addedamp)to an alkene no re~ction takes place because there is no electr~phile present tos~art a reactton by addmg to the nuCleophilic alkene The O-H bond of water are too strong

CH - CH = C~h + ~O----4 X

If however an acid (HeS040r HCI) is added to the solution a reaction will occur because the acidprovides an electrophile

Tle second order rale conslanl (M-Sec -) lor acid calalysed hydration at 25C is give n lor each 01 Ihe

fdllOwing akenesI H3CI

i C=CHz

Compound P Compound Q Compound R Compound S CompoundT

FRateconstant K K2 K3 K

1k Which is more reactive between (Z)-2 butene and (E)-2 butene towards hydration(A) (E)-2-butene (B) (Z)-2-butene(C) Both are equally reactive (D) Both are inert towards hydration

Sol B(E) isomer is more stable hence more free energy of hydration required

Correct relation between rate constants(A) K3 lt K1 lt Kelt ~ lt Ks(C) K3 lt K1 lt K2 lt Ks lt ~A(A) due to relative stability of carbcation formed

(B) K2 lt K1 lt K3 lt ~ lt Ks(D) K3 lt K1 lt K2 lt ~ lt Ks

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AITS-poundT -1I-Paper-2)-PCM(S )-JE E(Advanced)13 25xam- wAtX laquoUtt

xy planeyz plane

16 Amongthe given compounds Q R Sand T which has maximum number of planes of symmehy(A) Q (B) R~S ~T

Sol 0There are three planes of symmetry for compound T

HC i CHbullbull j

Xl plane _ YC

HC l CH3

1two structural iSQrnersl

~~2eq)

Compound (A) which can be distilled in high vacuum gives positive test with Br2H20 but does not reactwith Na metal in ethereal solution but compound (B) and(C) gives H2(g) with sodium metalCompound (A) undergoes following sequence of reactions -I C

LOAH )B

IBHsITHFH0210H

Products

_ve -poundk laquo C ~ - ve iodoform testi poundY r fr

17 Compound A should be

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

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~~_QraquoY Compound 8

I (A) kOHr 11~OHA

26 AITS-PT -11-( Paper -2)-PCM(S )-JEE(Advanced)13toon) 07

~ ~O ~)

HO - I

(8)Compound C is(A) OH

ISol A

Siolution for the Q No 17 to 19C1oH1602(A

Degree of unsaturalion = 10 + 1- 16 = 32

Compound does not react with Na metal so it does not contain acidic groups like - C- OH - OHo

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This section contains 3 questions Each question contains statements givenin two columns which have to be matched The statements in Column I arelabelled A B C and D while the statements in Column II are labelled p q rsand t Any given statement in Column I can have correct matching withONE OR MORE statement(s) in Column II The appropriate bubblescorresponding to the answers to these questions have to be darkened as illustrated in the following example

If the correct matches are A - P sand t B - q and r C - P and qi and D- sand t then the correct darkening of bubbles will look like the following

1 Match the following

A(fpoundj0C0B0~0Q)craquo~00Q)D~009QL)

(A)(B)(C)

Column-IHCOOHCH3COOH

( r-g~OH---- c

o~ ~CHj- -C-OH

Column-ll(p) Decarboxylates on heating(q) Reacts with Br2P(r) With Tollens reagent gives silver mirror

(s) Decarbonylates (removal of CO) on heating with H2S04

(t) Reduces HgCI2

Sol A ~ p r s t B -~ q C ~P q 0 ~ q

Match the followingColumn-I

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

Column-II(p) Treatment of NaN02 HCI gives N-nitroso

yellow liquid(q) Treatment of NaN02 HCI gives

diazonium chloride(r) Treatment of CH3 - I (excess) followed by

AgOH heat gives alkene

(s) Treatment of HCI heat gives non-volatilesalt

(t) Cannot form H-bond with each other

Sol A ~ r s B ~ P r s C ~ S t 0 ~ q S

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(A) PMMA(B) Orion(C) Bakelite(D) Dacron

-Column-l I Column-ll

I(p) Thermosetting polymer(q) Thermoplastic polymer

I(r) Addition polymer(s) Condensation polymer

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

PMMA Polymethylmethacrylate coacH I

iCH2-r] CH- nj

Orion Polyacrylonitnle CNiCH2-jCH3 n

Bakelite Phenol formaldehyde OH OH

CH2 CHz

Dacron Polyethylenetetraphthalale

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Mathematics PART -III

SECTION-A

This section contains 9 muhiple choice questions numbered 1 to 9 Each question has 4 choices (A) (B)(C) and (D) out of which ONLY ONE is correct

1 Normals PO PB and PC are drawn to the parabola y2 = 100x from the point p = (a 0) If thetriangle aBC is equilateral then possible value of a is

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

Equation of nonnal at B is y = -t1x + 2at1+ at~

y =2J3x+4-3a+24J3a

It passes through (a 0)

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

3 The maximum value odJx4- 3x2 - 6x + 13 - Jx4 + 5x2 +41 is

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

Le IJ(y - 2)2 + (x -3)2 - ~(y + 2)2 + x2 iHere y = x If P(x y) be any point on the parabola then IPA - PBI ABWhere A(3 2) BO -2)gt AB = 9 + 16 = 5

x2 y2If d be the length of perpendicular drawn from origin to any normal of the ellipse - + - = 1

25 16then d cannot exceed

1(A) 2 (B) 1

(C) 2 (D) none of theseBEquation of nonnal is 5x sec 8 ~ 4y cosec 8 = 9

d = 9 925sec2 e + 16cosec2e J25tan2 8 + 16cot2 0 + 41

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

gt 25tan2 e + 16cot2 (1 ~ 40 gt 25tan2 (1+ 16coe (1+ 41 ~ 81gt d ~ 1

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30 AllS-PT U-Paper2)-PCM(S)-JEE(Advanced)13

(B) 1(D) 2

If tan-1A=tan-1(A2 +n)+~ is satisfied by atleast one integral value of A in [0 2J Then the4

possible integral values of 0 isare(A) -1(C) 0A

0 = 1- ~ - A2 which gives 0 = -1 onlyA+1

(D) none of these

If from a point (1 0) two tangents are drawn on exactly one branch of the hyperbola ~ ~ ~ = 1

then 0 belongs to

(A) (-1 -~)(C) ( 11

BTwo tangents can be drawn if the point lies between theasymptotes OAand OB

xThe asymptotes are y = t-

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

1 a-cIn acute angle triangle ABC r = r+ r- - rl and B gt - Then -- completely lies in ~ 3 b

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

If the curve - + - = 1 and (x - 20)2 + (y - 20)2 = ~ intersect each other orthogonally at a400 100

point (20 cos e 10 sin G) then G 1

(A) cot 2 = J2(C) cot~ = 100

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AIlS-PT -1I-(Paper-2-PC M(S -JEE(Advanced13 31)- JYn xum YQuraquoraquogtYC bullbullbullbull

B(20 20) lies on x(20cos8) + y(10sinO) = 1

400 100

e 2 1 e 1- case 0 1=gt cos + Sin b = =gt Sin = --- =gt cot- = -2 2 2

x2 y2 - - 1An ellipse 2 + -2 = 1 and the hyperbola XL - yL intersect orthogonally It is given that-the

a b 22

eccentricity of the ellipse is reciprocal of that of hyperbola then a2 is equal tob

(A) ~ 2(C) ~

4BEllipse and hyperbola are confocal (t ae 0) = (t 1 0)

=gt a = 2 and e = ~12

equaticmof the ellipse is ~ + L = 12 1

9it _ 5~ ~

A tangent at the point P(20 25) to a parabola intersects its directrix at (-10 9) If the focus of theparabola lies on x-axis the numbers of such parabolas are(A) 1(C) infinite

-Sol BLet S= (a 0)As PSl as~gt 9 =-1

20 - a -(10+ a)=gta=5Let the slope of the directrix be mEquation of directrix is y - 9 = m(x + 10)Now 120m- 25 + 9 + 10ml = 225 + 625Jm2+1=gt 25m2

- 480 m - 297 = 0It gives two distinct values of mHence two parabolas

(8) 2(0) none of these

P(20 25)

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1- 32 AITS-PT -1I-Paper-2)-PCMS)-JEEAdvanced)13W)Y)iQttOOXOO fzrooowvetiXlt XXf)yart= XYbullbull(OO y -010-

Reasoning Type

This section contains 4 questions numbered 10 to 13 Each question contains STATEMENT-1 (Assertion)ahd STATEMENT-2 (Reason) Each question has 4 choices (A) (B) (C) and (D) out of which ONLYqNE is correct

1h STATEMENT 1 If CI with least absolute value is a root of the quadratic equation ax2 + bx - a == 0

where a gt Oand b lt Othen seccoc1(tanex-tan1(cota) =-1

- bSTATEMENT 2 For the quadratic equation ax + bx + c = 0 sum of the roots = -- and producta

cof roots = -

a(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(D) Statement 1 is false and Statement 2 is trueS The roots ofax2 + bx - a = 0 are negative reciprocals of each other aE (-10)Now tan(eosa) = -TI + coC1(tan a)

~sec[ cot-1(tancx)- tan1(cota)] secTl=-1

11 STATEMENT 1 The area of a triangle whose vertices are the incentr~ eireumcentre and centroidoithe triangle whose sides are 18 em 24 em and 30 em is equal to 3 em2 becauseSTATEMENT 2 Centroid divides each of the medians in the ratio 2 1(A) Both the statements are true and statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2 is nol the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false (D) Statement 1 is false and Statement 2 is true

Sol DThe triangle given is right angledTake a triangle with vertices (0 0) (18 D) (024)Centroid = (6 8)Circumeentre = (9 12)6 = rsgt r = 6So ineentre (6 6)

SO area of the required triangle = ~ x 2 x 3 = 2 em

12 1t 2(J 2 2 - 2 2STATEMENT 1 If 0 lt 0 lt - then ecosec sin 8 + esec t cos e gt e because2

STATEMENT 2 AM of positive numbers is always greater than or equal to their GM(A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(D) Statement 1 is false and Statement 2 is true

~eee~O~_elFmJfE Ltd FIITJEE House 29 4Kalu Sarai Sarvaprlya Villar bullbullew Delhi 110016 Ph 46106000 26569493 Fax 26513942

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AITS-PT -1I~Paper-2-PCM(S)-JeE(Advanced13caxumWUscet (cmraquoxtC~laquolaquoxt1 raquoyt

Sol AAM~GM

=gt ecoseclesin2 e + esece cos2 e ~ 2~eoosec2esec26 (cOS2 esin2 e)= 2~ellSinecose (cOS2 esin2e)= e2sin226sin28

Now consider f(x) == xe2x2 which c1ear1yis decreasing in (0 1) f(x) gt e2

13 STATEMENT 1 There exists atleast one solution of the equation

20[x2 - x ~ 5+ 10[ x - ~] == 1+ SO(cosx) (where [J denote greatest integer function)

STATEMENT 2 If x E R then x = [xJ + x where x denotes the fractional part of xbull (A) Both the statements are true and Statement 2 is correct explanation of Statement 1(B) aoth the Statements are true and Statement 2 is not the correct explanation of Statement 1(C) Statement 1 is true and Statement 2 is false(0) Statement 1 is false and Statement 2 is true

Sol DRHS is always an odd number So no solution

Comprehension Type ~ ~

i~This section contains 2 groupsibfquestions Each group has fmultiple choice question based~on aparagraph Each question has 4 choices (A) (B) (C) and (0) for its answer out of which ONLY ONE istorrect l t

Paragraph for Question Nos~ 14 to 16

A(am~ --2am)

(C) indined at ~6

oEquation of normal through A isy == mlx-2aml-a~

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

Normals are inclined at right angles

Read the following write up carefully and answer the following questionsA normal to the parabola at a point A on the parabola whose equation is yL = 2013x cuts the axis of x at

N AN is produced to the point B such that NB = AN Now answer the following questions2

If two more normals to the parabola yL = 2013x pass through B then they are

(A) coincident (B) inclined at ~4

(D) inclined at ~2

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If C is a point on the parabola l = 2013x such that 10CN == ~ (0 being origin) then projedion of

CN on x-axis is equal to

34 AITS-PT -IHPaper-2)-PC M(S)-JEE(Advanced)13 bull gtCb bullbullraquo))gtXOSrT

(A) 2013

(C) 4026A

If C ==(at2 2at) equation of CN is

y-2at= -~(x-at2)2

gtN==(at2+4a0)

Projedion of CN on x-axis = ON - OL = 4a

(B) 20132

(D) none of these

= 2013

16 If the nonnals through any point of the parabola y2 = 2013x cutthe line 2x - 2013 = 0 in poi1tswhose ordinates are in AP then the slopes of these normals are in(A) AP (B) GP(C) HP (D) none of these

S I th fl h 2013 th d 2013 3 0 vmg e equation 0 nonna Wit x = -2- e or mates are --4- m1

2013 3gt ---m24

2013 3gt ---m34As the ordinates are in AP 3 3 2 3 m1 +1Tl3 == m2

But ml + m3 = - m2Solving we get m~ ==m1m3

(B) the line y = x(D) the point (1 1)

(B) 2M-1(D) none of these

If (0 + f3f = 3 then (0 - (3)2 equals(A) 2J5 -1

(C) J1O-1BSol

Riead the following write up carefully and answer the following questionsThe ellipse whose equation is given by x2 + 4 = 4 is rotated about origin through a right angle in its own

Iplane The tangents at any point A on the ellipse before and after rotation intersect at P(o B)Nbw answer the following questions

1if The locus of P is symmetric about(A) origin(C) the line y = 2xA

1 Which of the following pairs of points lie on the locus of P(A) (0 3) and (0 - -J3) (B) (0 5) (0 -13)(C) (0 J5) and (1 1) (D) none of these

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AITS-PT -II-(Paper -2)-PCM(S)-JE EAdvanced)13raquoW iYtlF1W XYtCit iXOtYiXS N

17-19 Equation of any tangent before rotation is

XCOSa+ ysina = ~4cos2a + sin2 aEquation of the tangent after rotation is

J -xsina + ycosa == 14cos~ a + sin~ uSolving (1) and (2) we get

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

This section contains 3 questions Each question contains statements givenin two columns which have to be matched The statements in Column I arelabelled A B C and D while the statements in Column II are labelled p q rsand t Any given statement in Column I can have correct matching withONE OR MORE statement(s) in Column II The appropriate bubblescorresponding to the answers to these questions have to be darkened asillustrated in the following exampleIf the correct matches are A - p sand t B - q and r C - P and q and D - sand t then the correct darkening of bubbles will look like the following

A~00B0(V00c ~ltWgt000D~00()

~9Jtt~U (p) Straight line(A)

(C)(D)

Locus of centre of circles touching circle + - 4x - 4y= 0 internally and x2 + y2 - 6x - 6y + 17 = 0 externally isThe locus of the point (3h - 2 3k) where (h k) lies on the (q)circle x2 + ylt- 2x - 4y - 4 = 0 Locus ~f ce~tre of circle touching two circle XL + + 2x = I (r)o and Xi + t -6x + 5 = 0 externally is IThe extremities of a diagonal for a rectangle are (0 0) and I (s)(44) The locus of the extremities of the other diagonal is

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

2n 4n 6n 1998n -k () (p) 2(A) _ If cos--cos--cos-- cos-- = 2 then k + 1 IS1999 1999 1999 1999

divisible by

(B) If (1+tan10)(1+tan20) (1+lan450) = 2Jl then (n+1) is divisible (q) 3

by(C) The points (0 0) (a 11) and (b 37) are the vertices of an (r) 5

equilateral triangle then ab is divisible by

( 1 J (s) 7(D) tancos- sintan-I- is divisible by 42

Sol (A) ~ (p r) (8) ~ (p q) (C) ~ (q r s) (D)~ (p q s)

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36 AilS-PI --(Paper -2-PCM(S-JE E(Advanced )13

(A) Let a = ~ and P denote the desired product1999

Let Q = sinasin2a sin999a then2999pQ= (2sinacos a)(2sin2acos2a) (2sin999acos999a)

= sin2asin4a sin1998a= (sin2asin4a sin998a)[ -sin(2n -1000a)][ --sin(27t -1002a) J [- sin(21t -1998a)]

=sin2asin4a sin998asin999asin997a sina = QIt is easy to se that Q 0 hence the desired product is P =T999

~ k + 1=1000Hence P Rand T are correct options

(B) (1+ tankO)(1 + tan( 45 - k)O)= 1+[tankO + tan(45 - k)OJ + tanka tan(45 - k)o

=1tan45deg [1-tankO tan( 45 -kt]+ tanko tan(45 - kt = 1+tan45deg = 2

Hence (1 + tan10)( 1+ tan 44deg ) (1 +tan22deg )(1+ tan 23deg )(1 + tan 45deg )

~ n+1= 24Hence P Q and T are correct options

(C) Let 0 (00) A =(a 11) and B= (b 37) and M be the mid point of segment AS

(a+b ~M=1-241

Now OM l- MA and IOMl = J3IMAI-- a - b ----)3

Because AM=--i-13j if follow that OM=133i+~(a-b)j 2 2

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

a+b ~ a-b ~ --= 133 and -- = 83 2 2= a = 21)3 and b = 5-8 ~ ab= 315Hence Q Rand S are correct options

1 71 (n J 1 n(D) Because cos SlOB = - - 0 and tan - - e = -- for 0 lt 0 lt - we have for any xgt 0 2 2 tanO 2

t -1 -1 t l)1 t 1) 1ancos smtan x = an - - an x = -2 X

( 1 ~ tancos-1 sintan-1l

42) = 42

Hence P Q and S are correct optionsMatch the following

part of a rectangular hyperbolaa pair of lines

bullcO)9tfin~nbullbulla squarea circular arc

(p)(q)

Calufiiri~J ltiX + cosy ()

sinx + cos y + cos (xy) = ~2

secx + tany = 0 (r)

tanx2 + cor y = (s)2 I

(A) --t (q) (8) ~ (p) (C) ~ (r) (D)-~ (s)Use formula for atjdition of inverse trigonometric functions

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oo~ FIIJEE JEE[Advanced]2013

0-----ti--__--)

18------1- _

IA ----J8 i--------

oiDIA1

PART -I

ANSWERS HINTS amp SOLUTIONSPART TEST - II

PAPER-1

12 Dr- ~13 ~

14 I A15 8--

18 819

Physics

I FIITJff Ltd F1ITJEE House 29A Kalu -Sarai Sanapriya VlhaT New DeIhl] J0016 Ph 46106000 26569493 Tax 26513942~ u~bsitewwwfiifje-ecom

e IA t

B I--JI

IA II i~

12 Ie~_IA114 I B

115 I DI

116 i BI I~ Ie18 D

Chemistry PART - II

I~ __--J-~~--_ ~--------L-_~---- __~~_--~~ __~~ _

----------------------------__---------------------~~~~~~~~~e~_~~

C 0 0- wC A A

C 0 A-_-

0 C 0~ 0

(A) -) (r)

(B) -) (r)

(C) -) (p)

(0) -) (q)

(A) -) (p)

(B) -) (p q)(C) -) (s)(0) -) (r)

(A) -)(q f s)(B) -) (r)(C) -) (q)

I (0) -) (p)

QiN611

Physics

PART -I

SECTION-A

4R~ 4B(Jr

(D) 2n x 10-6sec

ISalIIIII

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

(C) aoRCo

The force2R

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

M((rlaquoHO lt )(

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

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e IA t

B I--JI

IA II i~

12 Ie~_IA114 I B

115 I DI

116 i BI I~ Ie18 D

Chemistry PART - II

I~ __--J-~~--_ ~--------L-_~---- __~~_--~~ __~~ _

----------------------------__---------------------~~~~~~~~~e~_~~

C 0 0- wC A A

C 0 A-_-

0 C 0~ 0

(A) -) (r)

(B) -) (r)

(C) -) (p)

(0) -) (q)

(A) -) (p)

(B) -) (p q)(C) -) (s)(0) -) (r)

(A) -)(q f s)(B) -) (r)(C) -) (q)

I (0) -) (p)

QiN611

Physics

PART -I

SECTION-A

4R~ 4B(Jr

(D) 2n x 10-6sec

ISalIIIII

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

(C) aoRCo

The force2R

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

M((rlaquoHO lt )(

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

C 0 0- wC A A

C 0 A-_-

0 C 0~ 0

(A) -) (r)

(B) -) (r)

(C) -) (p)

(0) -) (q)

(A) -) (p)

(B) -) (p q)(C) -) (s)(0) -) (r)

(A) -)(q f s)(B) -) (r)(C) -) (q)

I (0) -) (p)

QiN611

Physics

PART -I

SECTION-A

4R~ 4B(Jr

(D) 2n x 10-6sec

ISalIIIII

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

(C) aoRCo

The force2R

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

M((rlaquoHO lt )(

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Physics

PART -I

SECTION-A

4R~ 4B(Jr

(D) 2n x 10-6sec

ISalIIIII

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

(C) aoRCo

The force2R

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

M((rlaquoHO lt )(

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

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00000008

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(D) 2n x 10-6sec

ISalIIIII

(A) 1t x 10-6 sec (B) 2 x 10--6 sec4

(C) 2 x 10--6 sec2

(C) aoRCo

The force2R

o 2r2 0 4rL

~E~ ~E2ld03r 0 r2

M((rlaquoHO lt )(

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

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00000024

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00000029

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00000031

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00000042

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00000068

6yn x gt a )xxszSCO

(A) 2 (B) 3(C) 25 (0) 35

Sol AE x2 E 2

v = 2 mts

(f 0) x

(B) 4nR2KTP

(D) lIR2KT

(A) 2nR2KT

(C) nR2KTP

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(B) +0

(C) (D)

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Comprehension Type

Immersion heaters

Drain plug

(8) 57 x 10~kg(D) 48 x 103 kg

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

R B~0-1g

2Bv (B) 2coR2Bv

B2conR2d

1--------m

(0) B2poundoTIR

(B)(A)

9BconR2d1+-----

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

SECTION - 8

positive

_-IT17_-_

r --~--rr ~ i~---- ~ 2 ~-1l~ 71IL ~~

a- ~~~ ~

pOrT0I2 To T

Tlt= Tg PPo

(A) -) (P S t) (8) ~ (q) (C) - (q) (D) -) (q t)

(I) I Heat issystem

+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

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+++++++

-+~p p x

z-axis is along

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

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00000068

x~E =amposin(rot)

II (r)

r ~ St-NV( -

I -~ s~1VwJ

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Chemistry PART - II

SECTION-A

Et Et c=c H H

Et H c=c

Me Me c=c

AMe Me

bull-~1

(C) (D)

ir--I~CH3l~

(F) f)jC~CH~I -

F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

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00000006

00000007

00000008

00000009

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00000015

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F_--- CHO

I(( ---CH- C-OH- I

degIIHC-CH-C-OH- I

8rNHo -

RHbullC-CH-C-OH- I

H)CH amp(i) (ii) NO

r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

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00000031

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00000042

00000043

00000044

00000045

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r6l pKa=835

Ph-OH pKa = 998

H3C-C-OH pKa=476

X amp Yare

2 lsClpyriltjne

OTs OH

~_ ~ -6uOKHluOH) y

0s0 bull ~VOH

m-CPBAmCH-)

co 00~fOt-yoOyato~

Cb lsCtiP)

(ro~t~CD 0-8u--BuOH

OCH3 CH3 CH2-O-CH

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Reasoning Type

STATEMENT-1

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

STATEMENT-2

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Comprehension Type

Jleasant smell)

11CHoO-Na~

2 CH3 - Br(1eq)

E HO H0 ) F 1

0 )8+0ZnHzO

IsrI15

Compound C is(A)

ISFI16

o CHOII 1

~C-C~-C--C-O-C~CH3

Sol BI

CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

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00000044

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CH~o 0~ l

(DHC-C-C-H II IIo 0

(F)H~C-C-CH2ICH3

(H) CI-C-CHI 2

(I)HC-O-C-CH3 (J)

o 0II II

jaq H2S04

O H2C-C-OHO

O-CH2CH2CH

Compound C is

(A) Q-CHCH2CH2CH

o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

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o0-11I CH2CH2C-CH3

o-CH2-CH2-CH2-CH3

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

22ltmormon

p q r s

A)00Et9B0~00)c(00)D (~)0)0

Column-II(p) -OH

(q) -Br

(r) -O-R

(s) -Et

(t) -N02

II (p)i

Column-II

~~~XirOT n CD3

~CH(r) d-iHII (s)

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

+ CIIc

R1041+

(A) CgHg (p)

(B) C9H~ (q)

(C) C4H~ (r)

(D) C4H~ (s)(t)

(A) W8(B)

(C) 0 (D) rgt

[] []V0 e

IbullI

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

PART - III

SECTION -A

o-pound=-1gtp=qq

Now 5p + 15q = 40 gtp + 3q = 8 gtp = q = 2

1 mt2 503t-=-=-----1 -1 22013m

1 503t=gt m=--- m=--e 4026

4026 )13gt t = -1--

(A) 7 (B) 8(C)9 (0) 10

Sol Bi

sinl2x2 1=13

3=gt x2 = 2(4n+1) for n = 0 1 only

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(C) x2 + l -x + 2y == 0Sol C

(6) 4(x2 +y2)-4X+y 0

(D) none of these

50 radius s sJ2

(6) 128b2c

(D) _ 3a3_

(A) 3a3

(C)3a3

=gt DE = CE ~ ~4

1 2 A 2A= -Sin -sln-2 3 3

(A) (-11)

( 1 1 (6) 1--== -1

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

D Y C Y

Reasoning Type

( 2 )32(C) a t + 1eC

P(at2 2at) o( ~ _2a1t t

2 + 1)32

)t4 + t6 t2

1(B) al j1+e + ~1+e J j--z

(D) a-v1+ tt2

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

2 (a 2aLet A = (at 2at) B = 12 - -t )

bulllI

2xxIx) = -- =gt [xl = 1x+ x

b2=gt -=1

a=gt a(1 - e2) = 1 =gt 2a gt 2

I ctgtrrect

I C)x2+4l=12SOl B

As ID -LBC

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(B) rx-~Y +(y-3i = 19 2)x2 y2

(0) -+-- = 13 6

(0) none of these

(B) J5

(B) 4x - 3)3y = 12)3

(D) none of these

(A) xy = 4

(A) ~2c

(C)~-j32

p (0~I2

(8y)2 = 16(3 + x)(3 - x)

~ x2 + 4l = 9

(C) y2 = 12x

UWttrfPY X HlOO )

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

SECTION-B

(Matrix Type)

p q r s

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Sol (A) ~ (p) (B) ~ (p q) (e) ~(s) (D)~ (r)

~ Ap x AQ == 1a2

-2 41~3lalsin e-

1 1 5 -+-----=1 = a=- 5

a 254-ex 45 2 5 1

Fora = - 1 + ex = - = I = i~442

For ex = 5 t = i2

50 P = l ~ 11 R = (4 4) 5 = (4 -4) Q = Q == (~ - 11 54 ) 4 i

75 Area of Irep = -

4(D) If P(a12

2at) then

Q = (af -2at) ~ a = -at2

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(p) (-13)

(q) (-Y -3)v( 4x)

(s) (9 12)

(A) ~(q r s) (B) ~ (r) (e) ~ (q) (D)~ (p)

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

2 0 C B A

3 ABC A

~ A A C 0

5 A ADD

6 A A 0 C

7 B A A C

PAPER2

Physics

I (C) ~ (q)i (0) ~ (r)

PART -I

oooABAA

891011121314

171819

o(A) ~ (r)(B) ~ (p)(C) ~ (5)(0) ~ (q)(~) ~ (5)(B) ~ (p)(C) ~ (q)(0) ~ (r)(A) ~ (5)

3 I (B) ~ (r)(C) ~ (q)

~_-lJ0) ~ (p)

aoeMltt

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

o(A) --) (q r t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) --) (q)

(A) --) (rs)(B) --) (p rs)(C) --) (5 t)(D) --) (q s)

(A) --) (r 5)(B) --) (p r s)(C) --) (5 t)(D) --) (q 5)

(A) --) (q r t)(B) --) (r)(C) --) (p 5)(D) --) (q 5)

(A) --) (p r s t)(B) --) (q)(C) --) (p q)(D) -~ (q)

10(A) --) (qf t)(B) --) (r)(C) --) (p s)(D) --) (q s)

(A) --) (r s)(B) --) (p- f 5)(C) --) (5 t)(D) --) (q 5)

(A) --) (p r 5 t)(B) --) (q) (C) --) (p q)I (D) --) (q)

(A) --) (p r s t)(B) --) (q)

(C) --) (p q)(D) --) (q)

(A) --) (r s)(B) --) (p r s)(C) -~ (s t)(D) --) (q s)

(A) --) (qr t)(B) ~ (r)

I (C) --) (p s)I (D) --~(q s)

2 C3 0

5 B6 B7 C8 B9 A10 B

t 14t I 15

II 2iIr-III 3II

Ohemistry PART - II

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

10 8 D

11 0 A~~~

12 A 0-

13 018

----1_

14 0 iA

15 A iB

16 B 10I

17 A 10I

118 B IA19 0 IB

(A) - (q)(8) - (p)(C)- (r)

(D) - (5)

(A) - (s)

(8) - (q) I(C) - (r)rl(D) - (q)

(A) - (p r)(8) - (p q)

I (C) - (q r s) iI (D) - (p q s) I

IA-----i8

o(A) - (p r)(8) - (p q)(C) - (q r s)(D) - (p q s)

(A) - (q)

(8) - (p)

(e) - (r)

(D) - (s)

(A) - (s)(8) - (q)

(C) -(r)(D) - (q)

(A) - (s) (A) - (q)(B) ----)-(q) (8) - (p)

(C) -(r) (C) - (r)

(D) - (q) (D) - (s)

(A) - (q) (A) - (s)

(8) - (p) I (8) - (q)(C) - (r) I (C) - (r)

(D) - (s) _~) -~ (q)

QillN~m9Vgt1 C

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

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00000067

00000068

-PART -

SECTION-A

1 bull

~YSICS

Vosin(rot)

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

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00000049

00000050

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00000061

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00000063

00000064

00000065

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00000067

00000068

RBox 2 wL1 = 2 R2

Rz 4gt

sOx 1 + Box 2

For Box 3

=gt 2R ~ 2R2 = R + R2=gt R = 3R2

42R2 4

= I + 2 = Vo sin( (J) t)4R2

Phase angle is zero

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000049

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00000068

-11+41-3J3ilt E = J44

(A) -11 + 4 - 313k144

(B) -11 - 41 + 313k144

(C)-1i+41ffi

(D) -11 =1~J2

A- E E13 ~ E = -- i+--k - 2Ey3k+ 2Ej2 2

r bull

(A) 1(B) 35(C) 23(D) 12

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

700tt2N xt laquo00

_1_ [3q -+ 6q -+ ~ + 12q -I = 04mo 3a 3a 3a 4a J= q = -18q

~~q11= 9q

(8) 5 ms(D) 2 ms

~ v = --~ = ~ = 3ms1+sin~ 1+1

dt dt~

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

2 2B2(B) ~-q-

(A) ~ (r r

2 (q2B2r)(C) 1t --

0 A+ B ---------------e+ EAVA

(B) 2R(D) 2R In (1+ 12 )

B 1 dq

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

1UJ) Boa2

(A) _2

(C) 21tCl)Boa2

(B) 1tlJ)Boa2

(D) 1S1troBoa2

-S--VIm81V- 1m8

(C) 13 VIm8

(D) zero

IEI =V 1m4

P lt- - - ~~ 4J3r - - - - - - -3gt0--------------------__j9 Q

- __ EZ~OO E _- - 4 4 - r r - E O 3

I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

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00000029

00000030

00000031

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00000033

00000034

00000035

00000036

00000037

00000038

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00000040

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00000042

00000043

00000044

00000045

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I501112

50[111

STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

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00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000049

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STATEMENT-2

Comprehension Type

1 2 3 4

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

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00000006

00000007

00000008

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00000018

00000019

00000020

00000021

00000022

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00000024

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00000068

(A) 0+(EoA1V (B)(EoAI~+302 d4 d)42

(C) (COA)V_S

_ (coA)V 30q- - -+-d 4 2

(D) 302

q -q o

(A) (C~A ) V (B) 0 + C~A (~) 2t7

-~+2V--q-=vCo Co2

_ _ CoV _ (eoA l(V)~q---l-)-

xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000010

00000011

00000012

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00000014

00000015

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00000018

00000019

00000020

00000021

00000022

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xtoxrmraquoraquox

(B) 10QoC(0) 30iit

1 3(A) 20 (8) 2

(C) ~ (0) ~10 10

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

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00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000047

00000048

00000049

00000050

00000051

00000052

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00000061

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00000068

P = 7 x 105 N 012

iii 2 +Y_4m3

I SECTION ~B

(0) P(Nm1) (s) 15

5 3 x10

A- t isobaric B

process

(A) ~ (r) (B) ~ (p) (C) -) (s) (D) ~ (q)

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

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00000049

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00000067

00000068

~ __ 2_K_O 3_Ko__ ~ __ 6_K_o __ I

in steady

ltSH--

bull 2R ~ T bull

(r) IssoeI

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000014

00000015

00000016

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00000018

00000019

00000020

00000021

00000022

00000023

00000024

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00000026

00000027

00000028

00000029

00000030

00000031

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00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

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00000048

00000049

00000050

00000051

00000052

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00000068

16tttMvtN

l2lA bull

100degC 10degC

(A) ~ (s) (B) ~ (p) (e) ~ (q) (0) ~ (r)

C ~ ~~ V q2 V

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

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00000028

00000029

00000030

00000031

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00000038

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00000042

00000043

00000044

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00000061

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00000064

00000065

00000066

00000067

00000068

Find q2 x 100q1

Sol (A) -t (5) (8) -t (r) (e) -t(q) (D) -t (p)

(A) ~MR200 = ~MR2(1+ae)2oo5 5())= ())o (1 - 2aO)

[(r)- 000] x 100 = (2aO)IOO= 04

(B) P = aA6004

P2 = (crN2)f600A(52) = 54P

P2 - P1 x 100 = 25P1

(C) change is zero

4CV 7CV 3CV(D)q2 = -5- - 2 = 10

q2 x 100 = 60 q

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

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00000068

PART - n

5ECTION-A

(~ 11~

~r il~

W 4 3 2(B) 3 4 2~ 3 4 4(D) 4 3 0

00 01 0--V

~ __ ~~i

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

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00000058

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00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(A) 4 (8) 3(C) 2 (D) 1

ltgt=lt1Cl

CbOOCbH H

4 In the reaction

~CI-~PNH

Product P should be

C1S~-~

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

(B)9(D) 10

(C) HO~

(A)- 0HO~

cBr 8~ i Na-H O-H -----HO~~)

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

2 hydrolysIs

Br2ampq) ) Phv

2 hydrolysis

(X-(P)

(A) -o

rA-(CHOi()~ CH3

eOOH(t3-ketoacidj

qCOOOl

diI HSO~

EtC o~

ItoC-OEI

oEtO-C)7deg

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

bullbull

~ CHJCOOH + CO

2HOo(6)

(8irc duct) )

Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

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00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

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00000034

00000035

00000036

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00000038

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00000040

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00000042

00000043

00000044

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00000048

00000049

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Reasoning Type

cordgt Wetc

Ph-S CH2

bullbullbull _l

Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000048

00000049

00000050

00000051

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Comprehension Type

c~rrect

CH - CH = C~h + ~O----4 X

i C=CHz

xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

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00000014

00000015

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00000020

00000021

00000022

00000023

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00000040

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00000042

00000043

00000044

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xy planeyz plane

HC i CHbullbull j

Xl plane _ YC

HC l CH3

~~2eq)

LOAH )B

IBHsITHFH0210H

Products

(A) ~OH

Y---OHnHC--O~

~VAy=CHO~ -

OH iSol B

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000049

00000050

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00000068

I (A) kOHr 11~OHA

~ ~O ~)

HO - I

ISol A

(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

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00000026

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00000029

00000030

00000031

00000032

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00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

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00000051

00000052

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(A)(B)(C)

( r-g~OH---- c

o~ ~CHj- -C-OH

(t) Reduces HgCI2

(A) CH - CH2 - CH2 - NH2

H3C-N-CH3ICH3

Ph - NH2

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

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00000064

00000065

00000066

00000067

00000068

I (t) Brittle

Sol A --) q r t B --) r C --) p s 0 --)q S

iCH2-r] CH- nj

CH2 CHz

SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000010

00000011

00000012

00000013

00000014

00000015

00000016

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00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

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SECTION-A

(A) 150(C) 350C

IPBC = ~ gt t1 = 2-36

y =2J3x+4-3a+24J3a

0 = -23a + 4-2 (25)+ (243)25

0 = -V3CJ+ (4-3)(503)+ (24)3)503= a = 350

(B) 250(D) 450

W2 ~3~4 ~5

1(2)2 -2 12221V x - 2 + (x ~ 3) - jx + 2) + X I

1(A) 2 (B) 1

Now 25tan2 e + 16cOt2 e ~ J25 x 16 2

(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000010

00000011

00000012

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00000014

00000015

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00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

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00000031

00000032

00000033

00000034

00000035

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00000037

00000038

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00000040

00000041

00000042

00000043

00000044

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(B) 1(D) 2

(D) none of these

then 0 belongs to

(A) (-1 -~)(C) ( 11

(1 ( 1

C = 1 -I = D = 1 - J 2 2 gt

=oE ---l 2 2

(A) f ~) (B) (~ ~)34) 32

(C)r~~) (Dgtf~~l2 2 2 3

Br = r2 + r3- rl

1 11 n~ -=--+-----

s s-b s-c s-a

a-cEl~ 11 b 3)

o 1(B) cot- =-2 2

l)(D) cot- = 20

7x2 y2

(A) cot 2 = J2(C) cot~ = 100

400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

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00000047

00000048

00000049

00000050

00000051

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400 100

a b 22

(A) ~ 2(C) ~

=gt a = 2 and e = ~12

9it _ 5~ ~

P(20 25)

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

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00000063

00000064

00000065

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00000068

Reasoning Type

cof roots = -

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

00000017

00000018

00000019

00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

00000052

00000053

00000054

00000055

00000056

00000057

00000058

00000059

00000060

00000061

00000062

00000063

00000064

00000065

00000066

00000067

00000068

Sol AAM~GM

A(am~ --2am)

(C) indined at ~6

N(2(2013) + 2013m~ 0)If B = (h k) then k = 2013 ml

KBut mlm~m~ = --- =gt m~m- = -1~ ~ 2013 lt ~

(D) inclined at ~2

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

00000009

00000010

00000011

00000012

00000013

00000014

00000015

00000016

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00000018

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00000020

00000021

00000022

00000023

00000024

00000025

00000026

00000027

00000028

00000029

00000030

00000031

00000032

00000033

00000034

00000035

00000036

00000037

00000038

00000039

00000040

00000041

00000042

00000043

00000044

00000045

00000046

00000047

00000048

00000049

00000050

00000051

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00000053

00000054

00000055

00000056

00000057

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00000064

00000065

00000066

00000067

00000068

(A) 2013

(C) 4026A

y-2at= -~(x-at2)2

gtN==(at2+4a0)

(B) 20132

(D) none of these

= 2013

2013 3gt ---m24

(C) J1O-1BSol

(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000023

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00000032

00000033

00000034

00000035

00000036

00000037

00000038

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00000042

00000043

00000044

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(x2 +y2l =4(y+a)2 +(y-xi

or (y+x)2+(y-xf ~16(Y+X)2 +4(Y-X)2

0)=00(

SECTION-8

(Matrix Type)

A~00B0(V00c ~ltWgt000D~00()

(C)(D)

Circle

Hyperbola

Ellipse

Sol (A) ~ (5) (8) ~ (q) (C) ~ (r) (D)~ (q)

divisible by

(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

00000008

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00000014

00000015

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00000036

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00000042

00000043

00000044

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00000049

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(a+b ~M=1-241

a+b - J3 Hence ~-+24j=133i+-(a-b)j2 2

42) = 42

(p)(q)

secx + tany = 0 (r)

00000001

00000002

00000003

00000004

00000005

00000006

00000007

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Documents

Aits Part Test - Ii_qn & Sol