JEE Advanced 2013 Paper 1 Code 3 Final

7/29/2019 JEE Advanced 2013 Paper 1 Code 3 Final

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JEE Advanced 2013 Paper 1 (Code 3)

PARTI : PHYSICS

SECTION - 1: (Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B),(C) and (D) out of which ONLY ONE is correct.

1. A particle of mass m is projected from the ground with an initial speed u0 at an angle with the horizontal. At the highest point of its trajectory, it makes a completelyinelastic collision with another identical particle, which was thrown vertically upwardfrom the ground with the same initial speed u0. The angle that the composite systemmakes with the horizontal immediately after the collision is

(A)4

(B)4

(C)2

(D)2

Sol. (A)

Speed of first particle at the highest point, v1 = u0 cos

Speed of second particle at the highest point, v2 =20

u 2gH

Now,2 20u sinH

2g


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v2 =2 2

2 00

u sinu 2g

2g

= u0 cos

Final momentum = mu0 cos i + mu0 cos j

y o

x o

p u costan 1

p u cos

4

2. The image of an object, formed by a plano-convex lens at a distance of 8 m behindthe lens, is real and is one-third the size of the object. The wavelength of light inside

the lens is2

3times the wavelength in free space. The radius of the curved surface of

the lens is

(A) 1 m

(B) 2 m

(C) 3 m

(D) 6 m

Sol. (C)

air

med

fc 3

v f 2

Now, v 8m

1 vm u 24m3 u


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1 2

2

1

1

1

1 1 1

f v u

1 1 1 4

f 8 24 24

f 6m

1 1 11

f R R

R

1 11

f R

1 1 1

6 2 R

R 3m

3. The diameter of a cylinder is measured using a Vernier callipers with no zero error. Itis found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of themain scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th

division of the Vernier scale exactly coincides with one of the main scale divisions.The diameter of the cylinder is

(A) 5.112 cm

(B) 5.124 cm

(C) 5.136 cm

(D) 5.148 cm

Sol. (B)

1 MSD = 5.15 5.10 = 0.05 cm

1 VSD =2.45

0.049cm50

LC = 1 MSD 1 VSD = 0.001 cm

Reading = MSR + (LC x Vernier coincidence)

= 5.10 + (0.001 24)


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= 5.10 + 0.024

= 5.124 cm

4. The work done on a particle of mass m by a force,

3 / 2 3 / 2

2 2 2 2

yxK i j

x y x y

(K being a constant of appropriate dimensions), when the particle is taken from thepoint (a,0) to the point (0,a) along a circular path of radius a about the origin in thex-y plane is

(A)2K

a

(B)k

a

(C)K

2a

(D) 0

Sol. (D)

3/2 3 /22 2 2 2

3 /22 2

x yF K i j

x y x y

dW F.dr F. dx i dy j

xdx ydydW K

x y

Let x2 + y2 = a2

3

0 a

3a 0

2 2

3

KdW xdx ydy

a

KW xdx ydya

K a aW 0

2 2a

5. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to anend of another horizontal thin copper wire of length L and radius R. When the


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arrangement is stretched by applying forces at two ends, the ratio of the elongationin the thin wire to that in the thick wire is

(A) 0.25

(B) 0.50

(C) 2.00

(D) 4.00

Sol. (C)

Force applied is same.

2

2

21 1 2

2 2 1

2

2

1

F / A FLNow, Y

L /L R LFL

LR Y

L L R

L L R

2L R 1

L 2R 2

LElongation in thin wire2

Elongation in thick wire L

6. A ray of light travelling in the direction 1 i 3 j2

is incident on a plane mirror. After

reflection, it travels along the direction 1 i 3j2

. The angle of incidence is

(A) 30o

(B) 45o

(C) 60o

(D) 75o

Sol. (A)


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oi 3 j i 3 j

.2 2

cos 180 2i 3 j i 3 j

2 2

o

o

1 34cos21

1cos2

21

cos22

2 60

30

7. Two rectangular blocks, having identical dimensions, can be arranged either inconfiguration-I or in configuration-II as shown in the figure. One of the blocks hasthermal conductivity and the other 2 . The temperature difference between theends along the x-axis is the same in both the configurations. It takes 9 s to transporta certain amount of heat from the hot end to the cold end in the configuration-I. Thetime to transport the same amount of heat in the configuration-II is

(A) 2.0 s

(B) 3.0 s

(C) 4.5 s


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(D) 6.0 s

Sol. (A)

1

1 1

1

2

2

2 2

2

2

1

2

2

L L 3LCase 1: R

A 2 A 2 A

Q Tt R

Q 2 AT ------------(1)

t 3L

1 1 1 3 ACase 2:

R LL L

A 2 A

LR

3 A

Q T

t R

Q 3 AT ------------(2)

t L

Divide (1) by (2),

2 At 3L

3 At

L

t 2

9 9t 2sec

8. A pulse of light of duration 100 ns is absorbed completely by a small object initiallyat rest. Power of the pulse is 30 mW and the speed of light is 3 108 ms-1. The finalmomentum of the object is

(A) 17 10.3 10 kgms

(B) 17 11.0 10 kgms

(C) 17 13.0 10 kgms

(D) 17 19.0 10 kgms

Sol. (B)

Energy of the pulse, E = P x t = 30 x 10-3 x 100 x 10-9 = 3 x 10-9 J


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Since momentum is conserved,

Final momentum of the object = initial momentum of the pulse

917 1

8

E 3 10p 1 10 kgms

c 3 10

9. In the Young's double slit experiment using a monochromatic light of wavelength ,the path difference (in terms of an integer n) corresponding to any point having halfthe peak intensity is

(A) 2n 12

(B) 2n 14

(C) 2n 18

(D) 2n 116

Sol. (B)

max

max o

o o o o o

1I I

2

I 4I

1

I I 2 I I cos 4I2cos 0

n 2n 12 2

Path difference, x=2

2n 12 2

2n 14

10. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2:3.The ratio of their partial pressures, when enclosed in a vessel kept at a constanttemperature, is 4:3. The ratio of their densities is

(A) 1 : 4


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(B) 1 : 2

(C) 6 : 9

(D) 8 : 9

Sol. (D)

1 1 2

2 1 2

1

2

1

2

RTP

M

P M

P M

4 3

3 2

8

9

SECTION - 2 : (One or More Options Correct Type)

This section contains 5 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE OR MORE are correct.

11. Two non-conducting solid spheres of radii R and 2R, having uniform volume chargedensities 1 and 2 respectively, touch each other. The net electric field at a

distance 2R from the centre of the smaller sphere, along the line joining the centres

of the spheres, is zero. The ratio 1

2

can be

(A) -4

(B)32

25

(C)32

25

(D) 4

Sol. (B, D)


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1

P1

31

22

oo

1

2

2

P2

331 2

2 2o o

1

2

For internal point P ,

E 0

4R

R334 2R

4

For internal point P ,

E 0

4 4R 2R

3 3 04 2R 4 5R

32

25

12. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonicaccording to the equation, y(x, t) = (0.01 m) sin [(62.8 m-1) x] cos[(628 s-1)t].Assuming = 3.14, the correct statement(s) is (are)

(A) The number of nodes is 5

(B) The length of the string is 0.25 m

(C) The maximum displacement of the midpoint of the string, from its equilibriumposition is 0.01 m

(D) The fundamental frequency is 100 Hz

Sol. (B, C)


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No. of nodes = 5 + 1 = 6

y(x, t) = (0.01 m) sin [(62.8 m-1) x] cos[(628 s-1)t]

y(x,t) = 0.01m sin (20 x) cos(200 t)

k = 20 , =200

2 20.1m

k 20

5 5 0.1Length of the string, L 0.25m

2 2

The mid-point is antinode. Its maximum displacement = 0.01 m

v 200f 20Hz

2L k 2L 20 2 0.25

13. In the circuit shown in the figure, there are two parallel plate capacitors each ofcapacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then

released. The switch S2 is then pressed to charge the capacitor C2. After some time,S2 is released and then S3 is pressed. After some time,

(A) the charge on the upper plate of C1 is 2 CV0

(B) the charge on the upper plate of C1 is CV0

(C) the charge on the upper plate of C2 is 0

(D) the charge on the upper plate of C2 is -CVo


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Sol. (B, D)

When S1 is pressed and released



14. A particle of mass M and positive charge Q, moving with a constant velocity `1

1u 4i ms ,

enters a region of uniform static magnetic field, normal to the x-y

plane. The region of the magnetic field extends from x = 0 to x = L for all values ofy. After passing through this region, the particle emerges on the other side after 10

milliseconds with a velocity 12u 2 3i j ms .

The correct statement(s) is (are)

(A) The direction of the magnetic field is zdirection

(B) The direction of the magnetic field is +zdirection

(C) The magnitude of the magnetic field50 M

3Q

units

(D) The magnitude of the magnetic field is100 M

units3Q

Sol. (A, C)


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o

3

Mt

qB

Clearly 306

M 100 M 50 M6B 6q 3Qq 10 10

B must be in z direction.

15. A solid sphere of radius R and density is attached to one end of a mass-less springof force constant k. The other end of the spring is connected to another solid sphereof radius R and density 3 . The complete arrangement is placed in a liquid of density2 and is allowed to reach equilibrium. The correct statement(s) is (are)

(A) The net elongation of the spring is

34 R g

3k

(B) The net elongation of the spring is38 R g

3k

(C) The light sphere is partially submerged

(D) The light sphere is completely submerged

Sol. (A, D)


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Let the net elongation of the spring be x.

At equilibrium, for upper sphere

3 3

3

4 4kx R g R 2 g3 3

4 R gx

3k

33

33 3

4 16 R gTotal buoyant force = 2 R 2 g

3 3

4 4 16 R gWeight of the system = R g R 3 g

3 3 3

The system is completely submerged as total weight = total buoyant force.

SECTION - 3 : (Integer Value Correct Type)

This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9 (both inclusive).

16. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angularvelocity of 10 rad s1 about its own axis, which is vertical. Two uniform circular rings,each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the discin such a manner that they are touching each other along the axis of the disc and are

horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angularvelocity (in rad s-1) of the system is

Sol. (8)

By conservation of angular momentum,


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D D D ring

2D

2ring

2 2 2

D

2 2 2

1

I I 2I

1I MR

2

I 2MR (By theorem of parallel axis)

1 1

MR MR 2 2mr2 21 1

50 0.4 10 50 0.4 4 6.25 0.22 2

40 4

8 rad s

17. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratioof the slope of the stopping potential versus frequency plot for Silver to that ofSodium is

Sol. (1)

max

o

o

KE h

eV h

hV

e e

hSlope for both silver and sodium.

e

Ratio=1

18. A bob of mass m, suspended by a string of length l1, is given a minimum velocityrequired to complete a full circle in the vertical plane. At the highest point, it collideselastically with another bob of mass m suspended by a string of length l2, which isinitially at rest. Both the strings are mass-less and inextensible. If the second bob,after collision acquires the minimum speed required to complete a full circle in the

vertical plane, the ratio 1

2

l

lis

Sol. (5)

Speed of first bob at highest point = 1gl

For elastic collision between objects of same mass, velocities are exchanged

Speed of second bob = 1gl


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2

1 2

1

2

But minimum speed of second bob at the lowest point 5gl

gl 5gl

l5

l

19. A particle of mass 0.2 kg is moving in one dimension under a force that delivers aconstant power 0.5 W to the particle. If the initial speed (in ms-1) of the particle iszero, the speed (in ms-1) after 5 s is

Sol. (5)

f i

2

2

P t K K

10.5 5 0.2 v 0

2

2.5 0.1v

v 5 m / s

20. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103

disintegrations per second. Given that ln 2 = 0.693, the fraction of the initialnumber of nuclei (expressed in nearest integer percentage) that will decay in thefirst 80 s after preparation of the sample is

Sol. (4)

to

t

o

0.69380

T

o

0.69380

1386

o

0.04

o

N N e

N eN

Ne

N

Ne

N

Ne

N

No. of nuclei that will decay = No - N

% Fraction of nuclei that will decay in 80 s = o

o

N N100

N

0.04o

N1 100 1 e 100 1 1 0.04 100 4%

N


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PARTII: CHEMISTRY


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B),(C) and (D) out of which

ONLY ONE is correct.

21. The standard enthalpies of formation of CO2 (g), H2O(l) and glucose(s) at 25oC are -

400 kJ/mol, -300 kJ/mol and 1300 kJ/mol, respectively. The standard enthalpy ofcombustion per gram of glucose at 25oC is

(A) +2900 kJ

(B) 2900 kJ

(C) 16.11 kJ

(D) +16.11 kJ

Sol. (C)

6 12 6 2 2 2

oc r 2 r 2 r 6 12 6 r 2

oc

oc

oc

C H O s 6O g 6CO g 6H O l

H 6 H(CO ) 6 H(H O) H(C H O ) 6 H(O ,g)

H 6 400 6 300 1300 0

H 2900kJ /mol

2900H for onegramof glucose 16.11kJ /g180

22. KI in acetone, undergoes SN2 reaction with each P, Q, R and S. The rates of thereaction vary as

(A) P > Q > R > S

(B) S > P > R > Q

(C) P > R > Q > S

(D) R > P > S > Q

Sol. (B)


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23. The compound that does NOT liberate CO2 on treatment with aqueous sodiumbicarbonate solution, is

(A) Benzoic acid

(B) Benzenesulphonic acid

(C) Salicylic acid

(D) Carbolic acid (Phenol)

Sol. (D)

Phenols are weaker acids than carboxylic acids and hence do not liberate CO2 ontreatment with aq. NaHCO3 solution. Hence, benzoic acid, benzenesulphonic acid andsalicylic acid are more acidic than carbonic acid and will liberate CO2 with aq.NaHCO3 solution.

24. Consider the following complex ions, P, Q and R.

P = [FeF6]3-, Q = [V(H2O)6]

2+ and R = [Fe(H2O)6]2+

The correct order of the complex ions, according to their spin-only magnetic momentvalues (in B.M.) is

(A) R < Q < P

(B) Q < R < P

(C) R < P < Q

(D) Q < P < R

Sol. (B)

All the complexes have weak field ligands so there will be no pairing.

The electronic configuration of central metal ion in complex ions P, Q and R are


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(magneticmoment) n(n 2) B.M

No. of unpaired electrons (n) in P (Fe3+) = 5

No. of unpaired electrons (n) in Q (V2+) = 3

No. of unpaired electrons (n) in R (Fe2+) = 4

So, the correct order of spin only magnetic moment is Q < R < P

25. In the reaction,

P + Q R + S

The time taken for 75% reaction of P is twice the time taken for 50% reaction of P.The concentration of Q varies with reaction time as shown in the figure. The overallorder of the reaction is

(A) 2

(B) 3

(C) 0

(D) 1

Sol. (D)

From given data; t3/4 = 2 t1/2

So, the order of reaction with respect to P is one.


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From the graph, [Q] is linearly decreasing with time so the order of reaction withrespect to Q is zero. Therefore, overall order of the reaction is one.

26. Concentrated nitric acid, upon long standing, turns yellow-brown due to theformation of

(A) NO(B) NO2

(C) N2O

(D) N2O4

Sol. (B)

Conc. HNO3 slowly decomposes as

4HNO3 4NO2 + 2H2O + O2

It acquires yellow-brown colour due to the formation of NO2-

27. The arrangement of X- ions around A+ ion in solid AX is given in the figure (notdrawn to scale). If the radius of X- is 250 pm, the radius of A+ is

(A) 104 pm

(B) 125 pm

(C) 183 pm

(D) 57 pm

Sol. (A)

Cation A+ occupies octahedral voids formed by anions X- . The maximum radius ratiofor a cation to accommodate an octahedral void without distortion is 0.414. Radius ofanion X- is 250 pm.

A

x

A

R0.414

R

R 0.414 250 103.50 104pm

28. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is

(A) Fe (III)


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(B) Al (III)

(C) Mg (II)

(D) Zn (II)

Sol. (D)

Upon treatment with ammoniacal H2S, only Zn2+ ion gets precipitated as ZnS. Fe3+

ions and Al3+ ions also get precipitated as hydroxides but not as sulphides.

29. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25oC.For this process, the correct statement is

(A) The adsorption requires activation at 25oC

(B) The adsorption is accompanied by a decrease in enthalpy

(C) The adsorption increases with increase of temperature

(D) The adsorption is irreversible

Sol. (B)

The adsorption of methylene blue on activated charcoal is physiosorption. It isaccompanied by a decrease in enthalpy i.e. it is exothermic and does not haveenergy barrier.

30. Sulfide ores are common for the metals

(A) Ag, Cu and Pb

(B) Ag, Cu and Sn

(C) Ag, Mg and Pb

(D) Al, Cu and Pb

Sol. (A)

Silver, copper and lead are commonly found in earth's crust as Ag2S (silver glance),CuFeS2 (Copper pyrites) and PbS (Galena). The other metals Sn, Mg and Al exist as

SnO2, KCl.MgCl2.6H2O and Al2O3.xH2O respectively.

SECTION - 2: (One or More Options Correct Type)

This section contains 5 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE OR MORE are correct.

31. Benzene and naphthalene form an ideal solution at room temperature. For thisprocess, the true statement(s) is (are)

(A) G is positive


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(B) Ssystem is positive

(C) Ssurroundings = 0

(D) H = 0

Sol. (B, C, D)

Benzene and naphthalene form an ideal solution.

G ve for mixing

mixH 0 for ideal solution

systemS 0 since disorder increases and surroundingsS 0 because there is no

exchange of heat energy between system and surroundings.

32. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)

(A) 3 2 3 25 4Cr NH Cl Cl and Cr NH Cl Cl

(B) 3 2 3 24 2Co NH Cl and Pt NH H O Cl

(C)2 2

2 2 2 2CoBr Cl and PtBr Cl

(D) 3 3 33 3Pt NH NO Cl and Pt NH Cl Br

Sol. (B, D)

The pair of complex ions 3 2 3 24 2Co NH Cl and Pt NH H O Cl

shows geometrical

isomerism.

The pair of complexes 3 3 33 3Pt NH NO Cl and Pt NH Cl Br shows ionisation

isomerism.

The other pairs given do not have same type of isomerism.

33. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is1/100th of that of a strong acid (HX, 1M), at 25oC. The Ka of HA is

(A) 1 10-4

(B) 1 10-5

(C) 1 10-6

(D) 1 10-3

Sol. (A)


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Rate with respect to weak acid; R1

Rate with respect to strong acid; R2

R K[H ][ester]

1 2

Weak acid Strongacid

2a

2

4

1R R100

1 1H H M 0.01M C

100 100

0.01

K for weak acid C

1 0.01

1 10

34. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are

due to

(A) p (empty) and electron delocalisations

(B) * and electron delocalisations

(C) p (filled) and electron delocalisations

(D) p (filled) * and * electron delocalisations

Sol. (A)

In hyperconjugation, p (empty) electron delocalization for tert-butyl carbocation

takes place.

For 2-butene, electron delocalization takes place.


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35. Among P, Q, R and S, the aromatic compound(s) is/are

(A) P

(B) Q

(C) R

(D) S

Sol. (A, B, C, D)


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SECTION - 3: (Integer Value Correct Type)

This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9 (both inclusive).

36. The total number of lone-pairs of electrons in melamine is

Sol. Structure of melamine is


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So, melamine has six lone pair of electrons.

37. The total number of carboxylic acid groups in the product P is

Sol.

So, final product (P) has two carboxylic acids groups.

38. The atomic masses of 'He' and 'Ne' are 4 and 20 amu respectively. The value ofthe de Broglie wavelength of 'He' gas at -73oC is "M" times that of the de Brogliewavelength of 'Ne' at 727oC. 'M' is

Sol.

He Ne Ne

Ne He He

He Ne

h h

mv 2MK.E

SinceK.E T

hSo,

MT

For the twogasesHeandNe,

M T

M T

20 1000

4 200

5

39. EDTA4- is ethylenediaminetetraacetate ion. The total number of NCoO bondangles in [Co(EDTA)]1- complex ion is


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Sol.

Bond angles will be between N (I), N (II) with O (I), O (II), O (III) and O (IV).

So, total no. of N-Co-O bond angles are 8.

40. A tetrapeptide has -COOH group on alanine. This produces glycine (Gly), valine(Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For thistetrapeptide, the number of possible sequences (primary structures) with -NH2group attached to a chiral center is

Sol. According to question, C - Terminal must be alanine since -COOH group is intact onalanine so its NH2 must be participating in condensation and N - Terminal does havechiral carbon means it should not be glycine (optically inactive) so possiblesequences are:

Val Phe Gly Ala

Val Gly Phe Ala

Phe Val Gly Ala

Phe Gly Val Ala

So, answer is (4).


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PARTIII: MATHEMATICS


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B),(C) and (D) out of which ONLY ONE is correct.

41. The value of cot

23 n1

n 1 k 1cot 1 2k is

(A)23

25

(B)25

23

(C)23

24

(D)24

23

Sol. (B)

23 n1

n 1 k 1

231

n 1

231

n 1

231

n 1

231

n 1

231

n 1

cot 1 2k

cot 1 (2 4 6 ... 2n)

2n(n 1)cot 1

2

cot 1 n(n 1)

1tan

1 n(n 1)

(n 1) ntan

1 n(n 1)

231 1

n 1

1 1

23 n1 1 1

n 1 k 1

1

1

tan n 1 tan n

tan 24 tan 1

Now, we have :

cot cot 1 2k cot tan 24 tan 1

24 1cot tan

1 24.1

23cot tan

25

25

2

3


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42. Let PR 3i j 2k and SQ i 3j 4k

determine diagonals of a parallelogram

PQRS and PT i 2j 3k

be another vector. Then the volume of the parallelepiped

determined by the vectors PT,PQ and PS is

(A) 5(B) 20(C) 10(D) 30

Sol. (C)

PR a b 3i j 2k

SQ a b i 3j 4k

a 2i j 3k

b i 2j k

Volume of the parallelepiped determined by the vectors PT,PQ and PS is

given by

scalar triple product of2i j 3k , i 2j k , i 2j 3k .

Volume

2 1 3

1 2 1

1 2 3

= |2(6 2) + 1(3 1) 3(2 2)| = 10 cubic units

43. Let complex numbers 1and

lie on circles

2 2 2 22 2

o o o ox x y y r and x x y y 4r , respectively. If zo = xo +

iyo satisfies the equation2 2

o2 z r 2, then

(A)1

2

(B)1

2

(C)1

7

(D)1

3

Sol. (C)Equation of the two circles is(x xo)

2 + (y yo)2 = r2 ...(1)

(x xo)2 + (y yo)

2 = 4r2 ...(2)Equation of circle (1) and (2) can be written in complex from using zo = (xo + iyo) as

o oz z r and z z 2r


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Given,1

and

lies on the circles o oz z r and z z 4r respectively.

We have o o1

z r and z 2r

o

2o o

22 2o o o

o

2o o

2o o2 2

2 2o oo4 2 2

o o o

Consider z r

z z r

z z z r ... 3

1Now, consider z 4r

1 1z a 4r

z z 4r

z zz 4r

1 z z z

2 2 224r .... 4

2 22 2 2 2o o

22 2 2 2o

22 2 2o

222 2 2 2

o

22 2 2

2 2

On subtracting 3 from 4 , we get,

1 z z 4 1 r

1 z 1 4 1 r

1 1 z 4 1 r

r1 4 1 r Using given condition : 2 z r 2

2

r1 4 1 r

2

1 2 8

2

2

7 1

1

7

1

7

44. For a > b > c> 0, the distance between (1, 1) and the point of intersection of the

lines ax + by + c = 0 and bx + ay + c = 0 is less than 2 2. Then

(A) a + b c > 0(B) a b + c < 0(C) a b + c > 0


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(D) a + b c < 0Sol. (A)

ax + by + c = 0 ... (1)bx + ay + c = 0 ... (2)Using cross multiplication method,

2 2

x y 1

bc ac bc ac a b

2 2

2 2

2 2

(b a)c cx

a ba b

(b a)c cy

a ba b

According to the given information,

c c1 1 2 2

a b a b

c2 1 2 2

a b

a b c2

a b

a b c 2a 2b

a b c 0

45. Perpendiculars are drawn from points on the liney 1x 2 z

2 1 3

to the plane x+

y+ z= 3. The feet of perpendiculars lie on the line

(A)y 1x z 2

5 8 13

(B)y 1x z 2

2 3 5

(C)y 1x z 2

4 3 7

(D)y 1x z 2

2 7 5

Sol. (D)

x 2 y 1 z

2 1 3

Any point on this line is 2 2, 1, 3 This point will satisfy the equation of the given plane.

2 2 1 3 3

4 3 33

2

So, point of intersection is5 9

1, , C.2 2


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Let A (-2, -1, 0) be the point on the given line.Let B be the foot of perpendicular from A on the plane.

Direction ratio of AB is2 1

k.1 1 1

Any general point on line AB is (k 2, k 1, k).

This will satisfy the equation of the plane so, (k 2) + (k 1) + k = 3.k 2

Therefore,

, , 0,1,2 .

So, equation of the required line isx y 1 z 2

5 911 2

2 2

x y 1 z 2

7 51

2 2

x y 1 z 2

2 7 5

46. Four persons independently solve a certain problem correctly with probabilities

1 3 1 1, , ,2 4 4 8

. Then the probability that the problem is solved correctly by at least one

of them is

(A)235

256

(B)21

256

(C)3

256

(D)253

256

Sol. (A)

1 3 1 1

P A ,P B , P C , P D2 4 4 8

Required probability = 1- P(No one solves the problem correctly)


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1 P A P B P C P D1 1 3 7

1 . . .2 4 4 8

211

256

235

256

47. The area enclosed by the curves y = sin x + cos x and y = cos x sinx over the

interval 0,2

is

(A) 4 2 1

(B) 2 2 2 1

(C) 2 2 1

(D) 2 2 2 1Sol. (B)

The given curves are:

y = sin x + cos x = 2 sin x4

y = cos x sinx = 2 cos x4

Now, y = cos x sinx =

cosx sinx, x 0,

4

sin x cos x, x ,4 2


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/2 4 /2

0 0 4

/2 /2 / 4 / 4 /2 /2

0 0 0 0 / 4 /4

Area enclosed by the given curves

sin x cos x dx cos x sin x dx sin x cos x dx

cos x sinx sinx cos x cos x sinx

1 1 10 1 1 0 0 1 0 1

2 2 2

1

2

2 2 2 2

4 2 2

2 2 2 1 sq units

48. A curve passes through the point 1, .6

Let the slope of the curve at each point (x,

y) be y ysec , x 0.x x Then the equation of the curve is

(A)y 1

sin log xx 2

(B)y

cos ec log x 2x

(C)2y

sec log x 2x

(D)2y 1

cos log xx 2

Sol. (A)dy y ysec

dx x x

Put y vx

dvv x v sec v

dx

dxcosvdv

x

sinv logx c

It passes through 1,6

1 c2

Thus, the equation of the given curve isy 1

sin log xx 2

.


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49. Let1

f : ,1 R2

(the set of all real numbers) be a positive, non-constant and

differentiable function such that 1

f ' x 2f x and f 1.2

Then the value of

1

1/2f x dx lies in the interval

(A) 2e 1, 2e

(B) e 1,2e 1

(C)e 1

,e 12

(D)e 1

0,2

Sol. (D)

2x 2x

2x

2x

12

2x 2

2x 1

2x 1

2x 1

1 1 12x 1

1/2 1/2 1/2

2x 1

f '(x) 2f(x)

dy 2ydx

dye 2ye

dx

dye 0

dx

ye is decreasing function

1e f(x) e f

2

e f(x) e

f(x) e

Also, f(x) 0

0 f(x) e

0 dx f(x) dx e dx

e0 f(x) dx

2

11

1/2 1/2

1

1/2

e 1

0 f(x) dx 2

50. The number of points in 2, , for which x x sin x cos x 0, is

(A) 6(B) 4(C) 2(D) 0

Sol. (C)f(x) = x2 - xsin x- cos x


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f'(x) = 2x - x cos x - sin x + sin x= x(2 - cos x)

f(x) is increasing for x> 0f(x) is decreasing for x< 0f(0) = 1

f =

f = Thus, f(x) = 0 for exactly two values of x.

SECTION - 2: (One or More Options Correct Type)This section contains 5 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE OR MORE are correct.

51. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 :15 is converted into an open rectangular box by folding after removing squares ofequal area from all four corners. If the total area of removed squares is 100, theresulting box has maximum volume. Then the lengths of the sides of the rectangularsheet are(A) 24(B) 32(C) 45(D) 60

Sol. (A, C)Let length and breadth of the rectangular sheet be 8 and 15 respectively.

Let each side of the removed squares be x.Then, 4x2 = 100 x = 5 units


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3 2 2

2 2

2

2

V 8 2x 15 2x x

4x 46 x 120 x

dV12x 92 x 120

dx

The box has maximum value for x = 5.

dVThat is, 0 at x 5

dx

120 460 300 0

6 23 15 0

6 5 3 0

5or 3

6

Only = 3 is permissible.Thus, the lengths of sides of the rectangular sheet are 24 units and 45 units.

52. Let Then, Sn can take value(s)

(A) 1056(B) 1088(C) 1120(D) 1332

Sol. (A, D)

k k 14n 2 2

nk 1

S 1 k .


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k k 14n 2 2

nk 1

2 2 2 2 2 2n

2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2n

2 2 2 2

n

n

S 1 k

S 1 2 3 4 5 6

..... 4n 3 4n 2 4n 1 4n

S 3 1 4 2 7 5 8 6 11 9 12 10

.... 4n 1 4n 3 4n 4n 2

S 2 1 3 2 4 2 2 7 5 2 8 6

... 2 4n 1 4n 3 2 4n 4n 2

S

2 4n 4n 12 1 2 3 ... 4n

2

Now, consider option A.Let 4n(4n + 1) = 1056

2

2

4n n 264

4n n 264 0

n 8

Now, consider option B.Let 4n(4n + 1) = 1088

Now 1088 = 32 34

On comparing we get 4n (4n + 1) = 32 34, but such an integral value of n doesnot exist.

Now, consider option C.Let 4n(4n + 1) = 1120

Now 1120 = 32 35

On comparing we get 4n (4n + 1) = 32 35, but such an integral value of n doesnot exist.

Now, consider option D.

Let 4n(4n + 1) = 1332

4n2

+ n - 333 = 0 n = 9Hence, Sn can take values 1056 and 1332.

53. A line lpassing through the origin is perpendicular to the lines

1

2

l : 3 t i 1 2t j 4 2t k, t

l : 3 2s i 3 2s j 2 s k, s


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Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of

intersection ofland l1 is (are)

(A)7 7 5

, ,3 3 3

(B) 1, 1, 0 (C) 1,1,1

(D)7 7 8

, ,9 9 9

Sol. (B, D)

Equation of line l:x y z

a b c

Equation of the line l1,y 1x 3 z 4

t1 2 2

Equation of the line l2,y 3x 3 z 2

s

2 2 1

Direction ratio of line lis given by

i j k

1 2 2

2 2 1

2i 3j 2k

Equation of a line lisyx z

2 3 2

Point of intersection ofland l1,

2 3 t ... 1

3 2t 1 ... 2

On solving, 3 2 2 3 1 3 4 6 1

7 7

1

Point of intersection is (2, 3, 2)

So, 2 2 2

3 2s 2 3 2s 3 2 s 2 17

2 2 2

2

4s 4s 1 36 24s 4s s 17

9s 28s 20 0

10s 2,9

So, the required points are (1, 1, 0) and7 7 8

, ,9 9 9

.

54. Let f(x) = xsin x, x> 0. Then for all natural numbers n, f'(x) vanishes at

(A) a unique point in the interval1

n,n2


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(B) a unique point in the interval1

n ,n 12

(C) a unique point in the interval (n, n + 1)(D) two points in the interval (n, n + 1)

Sol. (B, C)f(x) = xsin xf'(x) = sin x x cos x 0

tan x x

The graph of tan x and x is as follows:

Clearly, from the graph, f'(x) has one root in1

n ,n 1 .2

Also, f'(x) has one root in

(n, n + 1).55. For 3 3 matrices Mand N, which of the following statement(s) is (are) NOT

correct?(A) NTMN is symmetric or skew symmetric, according as M is symmetric or skewsymmetric(B) MN NMis skew symmetric for all symmetric matrices Mand N(C) MNis symmetric for all symmetric matrices Mand N(D) (adj M) (adj N) = adj(MN) for all invertible matrices Mand N

Sol. (C, D)

T T

T T T T T TN MN N M N N M N

(A) IfMis skew symmetric,

TT TN MN N MN

Hence, skew symmetric.

IfMis symmetric, T

T TM MN N MN

Hence, symmetric. Option (A) is correct(B)


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T T T

T T T T

T T T T

MN NM MN NM

N M M N

M N N M

MN NM

(MN NM) is skew symmetric. Hence, option (B) is correct.

(C) T T TMN N M NM MN

Hence, option (C) is incorrect.(D) We know that: adj(MN) = adj(N) adjMHence, option (D) is incorrect.

SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9 (both inclusive).

56. A vertical line passing through the point (h, 0) intersects the ellipse 22 yx 14 3

at

the points Pand Q. Let the tangents to the ellipse at Pand Q meet at the point R. If

h = area of the triangle PQR,

max min1 2 1 21 /2 h 1 1 /2 h 18

h and h , then 85

Sol.2 2x y

S 14 3

Let the coordinates of P and Q be (h, ) and (h, ) respectively.

Now, P and Q lies on2 2

x y 14 3

2 2

22

2

h1

4 3

h3 1

4

h3 1

4

Thus, the coordinates of P are2h

h, 3 14

. Similarly, the coordinates of Q are

2hh, 3 1

4

.


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The equation of tangent at P isxh yk

14 3

and the equation of tangent at Q is

xh yk1

4 3 .

Thus, the point of intersection is R ,0h

.

2

3 /22

1 4Now, 2 h

2 h

h 43 1 h

4 h

4 h3

2 h

dNow, 0

dh

Therefore, is decreasing in1

,12

Thus, 11 15 45

2 8

2

1 2

9and 1

2

8 8 15 45 98 8 45 36 9

8 25 5

57. The coefficients of three consecutive terms of (1 + x)

n + 5

are in the ratio 5 : 10 : 14.Then n =Sol. Let three consecutive terms be r 1, r r 1t t , t

According to the given condition, we have:


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n 5 n 5 n 5r 1 r r 1C : C : C 5:10 :14

(n 5)! (n 5)!

10 14(n 5 r)!(r)! (n 5 r 1)!(r 1)!and

(n 5)! (n 5)!5 10

(n 5 r 1)!(r 1)! (n 5 r)!(r)!

(n 6 r)!(r 1)! 10 (n 5 r)!(r)! 14and(n 5 r)!(r)! 5 (n 4 r)!(r 1)! 10

n 6 r n 52 and

r

r 7

r 1 5

n r 2r 6 and 5n 5r 7r 18

7n r 2r 6 and n r r 1 5

5

Thus, we get,

72r 6 r 1 5

5

10r 30 7r 7 25

r 4n r 2r 6 n 4 2 n 6

58. Consider the set of eight vectors V= ai bj ck : a, b, c 1, 1 . Three non-coplanar vectors can be chosen from Vin 2p ways. Then p is

Sol. The eight vectors are:

1

2

3

4

5

6

7

8

v i j k

v i j k

v i j k

v i j k

v i j k

v i j k

v i j k

v i j k

Consider the vectors (v1, v5), (v2, v6), (v3, v7) and (v4, v8) in pair. If we take any oneof the four pairs consisting two vectors and a third vector out of reaming six vectorsthen these three vectors will be coplanar. Thus for each pair we have 6 options.Thus, the number of coplanar vectors = 6 + 6 + 6 + 6 = 24Also, out of 8 vectors, 3 vectors can be chosen in 8C3 = 56 ways

Number of non-coplanar vectors = 556 24 32 2 Thus, p = 5

59. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is , only E2 occurs is and only E3 occurs is . Let the probability p that none of

events E1, E2 or E3 occurs satisfy the equations 2 p and 3 p 2 .


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All the given probabilities are assumed to lie in the interval (0, 1). Then

1

3

Pr obability of occurrence of E

Pr obability of occurrence of E

Sol. We have:

2 31

1 32

1 2 3

1 2 3

P E P E P E ... i

P E P E P E ... ii

P E P E P E ... iii

P E P E P E ... iv

Now, from the given conditions for p, we have:


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60. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards areremoved from the pack and the sum of the numbers on the remaining cards is 1224.If the smaller of the numbers on the removed cards is k, then k 20 =

Sol. There are n cards numbered from 1 to n.

Sum of n cards=n(n 1)

2

If two consecutive cards numbered k and k+1 are removed then

S' =

n(n 1)

k (k 1) 1224 (given)2

2

n(n 1)1225 2k

2

n n 2450 4k 0

Now, n must be an integer. Thus, D must be a perfect square.


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2

2

2

2

2 2

Let D

Then,

1 4( 2450 4k)

1 16k 9800

16k 9801

16k 99

( 99)( 99)k

16

Now, k must be a positive integer which is possible when 101

(101 99)(101 99) 400k 25

16 16

Hence, k 20 25 20 5

The smallest value ofn for which

n n 11224

2n n 1 2448

n 49

For n 50

n n 11275

2

So, k k 1 1275 1275 51

k 25

k 20 5

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JEE Advanced 2013 Paper 1 Code 3 Final