Tes Pttern Eglish REVIEW TEST CLAS ONT PROGRAMME ... - ALLEN

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are toldto do so.

2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

3. Blank spaces are provided within this booklet for rough work.

4. Write your name, form number and sign in the space provided on the back cover of thisbooklet.

5. After breaking the seal of the booklet, verify that the booklet contains 28 pages andthat all the 18 questions in each subject and along with the options are legible. If not,contact the invigilator for replacement of the booklet.

6. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET :

7. The ORS will be collected by the invigilator at the end of the examination.

8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

9. Write your name, form number and sign with pen in the space provided for this purposeon the ORS. Do not write any of these details anywhere else on the ORS. Darkenthe appropriate bubble under each digit of your form number.

DARKENING THE BUBBLES ON THE ORS :

10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

11. Darken the bubble COMPLETELY.

12. The correct way of darkening a bubble is as :

13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correctway.

14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to eraseor "un-darken" a darkened bubble.

15. Take g = 10 m/s2 unless otherwise stated.

Please see the last page of this booklet for rest of the instructions

Time : 3 Hours Maximum Marks : 204

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PAPER-2

Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

SAMPLE PAPER-1

JEE(Advanced)

REVIEW TESTEngl

ish

E-2/28

Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-2

Space for Rough Work

SOME USEFUL CONSTANTSAtomic No. : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

ALLEN

SAMPLE PAPER-1/Paper-2ALLEN

E-3/28


BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART-1 : PHYSICSSECTION–I(i) : (Maximum Marks : 12)

� This section contains FOUR questions.� Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options

is correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

1. The operating temperature of a tungsten filament in an incandescent lamp is 2450 K and itsemissivity is 0.30. The surface area of the filament of a 25 watt lamp is :(Take : s = 5.67 × 10–8 S.I. unit)(A) 0.1 cm2 (B) 0.2 cm2 (C) 0.3 cm2 (D) 0.4 cm2

2. Two long wires are placed on a pair of fixed parallel rails perpendicular to the wires.The spacing d between the rails is large compared with x, the distance between the wires.Both wires and rails are made of a material of resistance r per unit length. A magnetic fluxdensity B is applied perpendicular to the rectangle so formed, and one wire is moved alongthe rails with a uniform speed v while the other is held stationary. It is seen that the forcerequired to hold the stationary wire vanishes at a particular value of x. Find x.

(A) 0v4m

pr (B) 0v2mpr (C) 02 vm

pr (D) 0v3mpr

Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-2ALLEN

E-4/28


3. What horizontal force F must be applied to block (figure), so that it moves down the inclinedplane with constant velocity? Mass of the block m = 2 kg, the coefficient of friction between theblock and the surface of the plane is 0.2; plane forms angle of 45° to horizontal.

Fm

a

(A) 20 N (B) 13.33 N (C) 8 2N (D) 4 2N

4. An athelete runs up and projects Javelin at a certain angle to the horizontal. If the speed ofthe Javelin relative to the athelete is equal to the speed of the athelete at the time of projection,what should be the angle at which he should throw relative to himself so that the range ismaximum? The height of the athelete can be considered negligible.(A) 45° (B) 30° (C) 60° (D) 37°


E-5/28


SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these

four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

both of which are correct options.Partial Marks : +1 If two or more options are correct but ONLY one option is chosen

and it is a correct option.Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).Negative Marks : –1 In all other cases.

� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correct option(s)will result in –1 marks.

5. A part of radiation spectrum corresponding to Lyman series from a hydrogen like atom is

shown here. One of the lines is missing. (Take : R = 7 11610 m

15-´ )

1254 nm

25 nm

(A) The atom is helium

(B) The missing line is 3375

nm128

(C) The spectrum shown here is absorption spectrum(D) Some atoms are excited to the third excited state.


E-6/28

6. Two sources are connected in the chain, as shown in Figure. Determine the potential difference V1between points A and B with an open external circuit, if e1 = 5 V, r1 = 1 ohm, e2 = 3V,r2 = 2 ohms. What will be the potential difference V2 between points A and B if you close thekey K ? Resistance R = 5 ohms. Also find the current through the batteries after closing theswitch.

A Be1 1, r

e2 2, rK

R

(A) V1 = 4.3 V approx(B) V2 = 3.8 V approx(C) Current through battery e1 = 1.2 A approx.(D) Current through battery e2 = 0.4 A approx

7. Two homogenous spheres of mass 1 kg each, radius 10 cm are released from same height on aninclined plane of inclination 37°. The coefficient of friction for the first sphere is 0.1 and for thesecond sphere is 0.3

37°

(A) The ratio of time taken by first and second to reach the bottom is 7591 .

(B) The ratio of translational kinetic energy of first and second when they reach the bottom is 1:1

(C) The ratio of speed of first and second when they reach the bottom is 9175 .

(D) The ratio of angular velocity of first and second about their center when they reach the

bottom is 9175 .



E-7/28

8. A fixed long ideal solenoid is encircled by a coaxial circular loop of radius R greater thanradius of the solenoid. A current I is passed through the circular loop. In which of the followingcases, no emf is induced in the solenoid ?

y

x

z

(A) The current in the loop is held constant and the loop is stationary.(B) The current in the loop is I = I0 sinwt and the loop is stationary.(C) The current in the loop is constant and the loop performs angular SHM about its diameter

(y axis) (but does not reach the solenoid surface)(D) The current in the loop is constant and the loop performs SHM about its mean position in

a direction perpendicular to the axis (but does not cross the solenoid)9. A vertically hanging rod of mass 5 kg is suspended from a hinge at A. A massless inextensible

cord is attached to the midpoint of the rod and is pulled at an angle of 37 degree to the vertical asshown. At the other end, a mass of 20 kg is hanging. Find F1 and F2 applied to keep the system inequilibrium, tension T in the cord and force F at the hinge.

20kgF1

F2

A

37°

(A) F1 = 75 N (B)F2 = 200 N (C) T = 250 N (D) F = 25 10N



E-8/28

10. A Rod of length L and mass m is suspended from the ceiling using two cords of same length L.It can be oscillated in two ways:(1) perpendicular to the plane of paper and (2) in the plane ofthe paper. The cords are massless and inextensible. Find the time period in each case.

LL

L

(A) in way (1) = p3 L

T 22 g

(B) in way (2) = p3 L

T 22 g

(C) in way (1) 5LT 2

3 3g= p (D) in way (2) 5L

T 23 3g

= p

11. In an infinite flat layer of thickness of h the spherical cavity in diameter of h (figure) is cut out.The layer is uniformly charged with volume charge density of r.

A

B

(A) The electric field at A is 0

h

6

re towards right.

(B) The electric field at B is 0

h

3

re in the upwards direction

(C) If the potential at the center of the cavity is zero, the potential at A is 2

0

h12r

e

(D) If the potential at the center of the cavity is zero, the potential at B is 2

0

13 h12

re



E-9/28

12. A triangular prism of sides 4m × 3m × 1m is completely filled with water. If we neglect theatmospheric pressure, the force by water (density = 1000 kg/m3) on

A

B

D

E

F

C3m

4m

1m

(A) On surface ABCD is zero(B) On surface CDEF is 2 × 104 N(C) On surface ADE is 8 × 104 N(D) On surface ABEF is 10 × 104 N



E-10/28

SECTION-II : (Maximum Marks: 24)� This section contains SIX questions.� The answer to each question is a NUMERICAL VALUE.� For each question, enter the correct numerical value (in decimal notation, truncated/rounded-

off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is 11.36777.....then both 11.36 and 11.37 will be correct) by darken the corresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluated according to the following marking scheme:Full Marks : +4 If ONLY the correct numerical value is entered as answer.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

1. The external and internal radius of a hollow sphere are measured to be (4.20 ± 0.01) cm and(3.90 ± 0.01) cm. Maximum percentage error in the measurement of volume will be.

2. A ring of radius R and mass m is uniformly charged with charge Q. It is rolling purely asshown. A uniform magnetic field of magnitude B is present perpendicular to plane of the ring. Ifthe normal force exerted by the ground is lmg, find l. Take QvB = mg.

+ + ++

++

+++

+++

v



E-11/28

3. Refer to the circuit given below. Initially the switch S is in position–1 for 0.2 ln 2 s. Then theswitch is changed to position–2. After a time t (measured from the change over of the switch) the

voltage across 5 kW resistance is found to be 35

volt. Then, find the value of t (in milli sec)

10V 2

1S

5kW

10kW

20 Fm

4. In the figure shown below, the maximum possible unknown resistance (X), that can be measuredby the post office box are Xmax is given by R × 105 W, then R is : (In this experiment, we take outonly one plug in arm AB and only one plug in arm BC, but in arm AD we can take out manyplugs):

G+–

E

1000 100 10 B10 100

D

A

500200200100 X

unknownresistance

QP

R

C1000

5000 2000 1000



E-12/28

5. What is the difference in liquid levels (in mm) of the two intercommunicating capillarieswith diameters 2 mm and 1mm ? Liquid surface tension is 0.07 N/m. Meniscus contact anglesare zero for both the capillaries. Fluid density is 1000 kg/m3.

6. Proton separation energy is the amount of energy we must supply to remove a proton from a

nucleus. Find the proton separation energy of He42 (in MeV) . The mass of

( )Hm 11 = 1.007825 amu

( )Hm 31 = 3.016049 amu

( )Hem 42 = 4.002603 amu

(Take : 1amu = 931 MeV).



E-13/28


PART-2 : CHEMISTRYSECTION–I(i) : (Maximum Marks : 12)




1. A vessel contains He gas at 20 atm and CH4 gas at 'P' atm. When a pin-hole is made in the

vessel, the gases coming out initially contains 80% He gas, by mole. The value of 'P' is(A) 1 atm (B) 40 atm (C) 2.5 atm (D) 10 atm

2.

OH

O

dil.NaOH¾¾¾¾¾®

Major form present in reaction mixture is

(A)

OH

O(B)

O

O

(C)

O

O(D)

O

C – ONa

O


E-14/28


3. Which of the following reaction gives carboxylic acid as final product

(A) O

O

O 3(i)H O(ii)Δ

Å¾¾¾¾® (B) CH – CH – C – H3 2

O mcpba

Δ¾¾¾¾®

(C) O O

OEt3(i)H O

(ii)Δ

Å¾¾¾¾® (D)

32

–

(i) O(ii) Zn,H O(iii)O H,Δ

¾¾¾¾¾®

4. Which is NOT produce in given reaction change?

Sodium-amalgam water¾¾¾¾®

(A) H2 (B) NaOH (C) Hg (D) O2


E-15/28


SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:




� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –1 marks.

5. Select the correct statements -(A) More solute is adsorbed on activated charcoal when solution is diluted(B) All colloidal solutions show Tyndall effect significantly(C) Dialysis may be used to coagulate the sol.

(D) Lyophobic sols can be coagulated easily compared to the lyophilic sols.


E-16/28


6. In a voltmeter, mass of a metal deposited in 30 seconds in 200 gms. Analyse the currentv/s time graph shown below and identify the correct statement(s):

i

100 mA

10sec 20sec 30sec

t(A) Electrochemical equivalent of the metal is 100 g/coulomb(B) 33.33 grams got discharged in the first 10 seconds.(C) A constant current of 66.66 mA would also discharge approximately the same amount in

same time(D) 100 gms of metal got discharged in the first 15 seconds

7. The reaction : A(s) �� 2B(g) is in equilibrium at 4 atm and 27°C. If the volume of systemis increased at constant temperature, then at new equilibrium, relative to initialequilibrium :-(A) Moles of A(s) decreases (B) Moles of B(g) increases

(C) Molar concentration of A(s) decreases (D) Molar concentration of B(g) decreases

8. Which of the following is classified into free radical substitution reaction

(A) R – C – OAgO

2 4Br ,CClΔ¾¾¾¾¾® (B)

2 2

HBrR O ,hν¾¾¾¾¾®

(C) NBSΔ¾¾¾® (D) 2

4

BrCCl¾¾¾®


E-17/28


9. Which of the following is/are used for preparation of aldehyde?(A) Stephen reduction(B) Rosenmund reduction(C) Clemmensen reduction(D) Reaction of 1-propanol by cromic anhydride in glacial acetic acid.

10. Out of given select correct order(s) :-(A) F > N > C > B > Si (Non metallic character)(B) F2 > Cl2 > O2 > N2 (Chemical reactivity in term of oxidising property)(C) K > Mg > Al > B (Metallic character)(D) Ne > F > O > N (First ionisaton energy)

11. Which is/are CORRECTLY represented with the explanation for colour of given species?(A) FeSO4.7H2O Þ t2g to eg transition(B) Br2 = s* to p* transition(C) KMnO4 = L ® M transition(D) Balmer series emission of H atom Þ De-excitation of electron from n = 3,4...... to n = 2

12. Which of the following is/are water insoluble?(A) Rhombic sulphur (B) Hydrogen bromide(C) Beryte salt (D) Ferric chloride


E-18/28



off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubblesin the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. Glutathione ( molecular weight = 307 gm/mol) is important antioxidant preventing damageto cellular components from free radicals, where it dimerises as

HS

NH

NH2

HO

O O

NH

HO

O

O

¾¾® OHN

HOO

SS

NHNH2

HO

O O

NH

HO

OO

HNH2N

OH

OO

Calculate gm equivalent weight of glutathione.2. A solution is prepared by mixing 1 M HA (pK

a = 4.2) and 1 M NaA. If pH of solution is 4.5. Find

the ratio of volume of NaA to HA added. (take : log 2 = 0.3).


E-19/28


3. How many compounds give yellow liquid with NaNO2 + HCl

(i) CH3 – CH2 – NH2 (ii) N – CH3

H(iii) NH2 (iv) N

H

(v) CH – N – CH3 3

CH3(vi) CH – – N – CH3 3CH2

H

(vii) CH – – NH – CH3 3C

O

(viii) CH – – NH3 2C

O

4. How many total synthetic condensation polymers are :(i) Glyptal (ii) PVC (iii) Cellulose (iv) Dacron(v) Orlon (vi) Nylon-66 (vii) Bakelite(viii) Urea formaldehyde resin

5. Most common oxidation number of 3d series metals is +2 but Sc+2 is virtually unknown.Find sum of 3d electrons in all the known M+2 ions of 3d-series.

6. In the three dimensional structure of a chiral complex compound [Co(Co(NH3)4(OH)2)3]Br6.Find total number of Co–N & Co–O linkages.


E-20/28


PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 12)




1. The sum of maximum & minimum values of expression y = 10sin2x + 6sinx cosx + 14 cos2x is-(A) 12 (B) 24 (C) 36 (D) 48

2. Number of integral solution(s) of the inequality 2sin2x – 5sinx + 2 > 0 in x Î [0,2p], is-(A) 3 (B) 4 (C) 5 (D) 6

3. Absolute value of slope of a line, common tangent to both the curves given by y = x2 andx2 + y + 1 = 0 will be -(A) 5 (B) 2 (C) 3 (D) 2

4. Complex numbers z1 & z2 are such that |z1| = 3, |z2| = 5 & |z1 + z2| = 7. The value of 6162

zarg

zmay be-

(A) p (B) 2

3

p(C)

3

p(D) 2p


E-21/28


SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:




� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –1 marks.

5. Let k Î Q and ƒ(x) = x2 – (3 – 2k)x + 2 – 4k, then identify the correct statement-(A) roots of ƒ(x) = 0 are rational

(B) if roots of ƒ(x) = 0 are of opposite sign, then 1k

2>

(C) if 1k

2£ , then roots of ƒ(x) = 0 are always of same sign.

(D) if exactly one root of ƒ(x) = 0 is greater than 2, then greatest integral value of 'k' is –1.


E-22/28


6. 7 points are given in a plane (as shown in adjacent figure), then using these 7 points -1

2

34 5 6 7

(A) Number of lines that can be formed is 10.(B) Number of triangles that can be formed is 24.(C) Number of quadrilateral that can be formed is 6.(D) Maximum number of circles that can be formed is 24, where each circle contains atleast three

of 7 points.7. There are 3 bags A, B & C. Bag A contains 1 Red & 2 Green balls, bag B contains 2 Red &

1 Green balls and bag C contains only one green ball. One ball is drawn from bag A & put intobag B then one ball is drawn from B & put into bag C & finally one ball is drawn from bag C &put into bag A. When this operation is completed, probability that bag A contains 2 Red &1 Green balls, is -

(A) 1

4(B) 1

2(C)

1

3(D)

1

68. If A and B are square matrices of order of 3 such that AB = aA + bB, where a, b Î R – {0}, then

the correct statements is/are.(A) A and B cummute (B) A – bI is invertible(C) B – aI is invertible (D) A = B


E-23/28

9. Function ƒ(x) satisfies the relation ƒ(xy) = ƒ(x).ƒ(y) for all positive numbers x,y and ƒ(2) = 10,then-(ƒ–1(x) represent inverse of ƒ(x))

(A) ( )4

r

r 0

ƒ 2 10000=

=å (B) ( )4

r

r 0

ƒ 2 11111=

=å

(C) ( )4

1 r

r 0

ƒ 10 31-

=

=å (D) ( )4

1 r

r 0

ƒ 10 32-

=

=å

10. If 21

x 2 21 cosx

x 0lim(5 x ) L.e-

®+ = , then the value of

1t t t t t 2t

tlim(1 2 3 ......... 25 L )®¥

+ + + + + is greater than

(A) 1 (B) 3 (C) 5 (D) 30Space for Rough Work


E-24/28

11. Let ƒ(x) is a continuous function in [a, b] such that ƒ(a) ¹ ƒ(b) and ƒ(a), ƒ(b) > 0, then which ofthe following is CORRECT ?

(A) If ƒ(c) =ƒ(a) ƒ(b)

2

+, then there always exist at least one 'c' in (a, b)

(B) If ƒ(c) ƒ(a)ƒ(b)= , then there always exist at least one 'c' in (a, b)(C) If 5ƒ(c) 3ƒ(a) 2ƒ(b)= + , then there always exist at least one 'c' in (a, b)

(D) If 2ƒ(c) 3ƒ(a) ƒ(b)= - , then there always exists at least one 'c' in (a, b)

12. If ( )3xP e cos 2013x dx= ò and ( )3xQ e sin 2013x dx= ò (where P & Q both have integration

constant zero) & ƒ(x) = ln((P2 + Q2) (32 + 20132)) then(A) ƒ(1) + ƒ(2) = 18 (B) ƒ'(1) = ƒ'(–1)(C) ƒ(x) is an odd degree polynomial (D) ƒ(x) is symmetric about x = 0



E-25/28


off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubblesin the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. If P1, P2, P3 are the points on ellipse 3x2 + y2 – 12 = 0 and P1', P2', P3' are their correspondingpoints on the auxillary circle, then the area of triangle P1' P2' P3' is l times the area of triangleP1P2P3, then l2 is

2. If the family of integral curves of the differential equation 3dyx y x

dx+ = is cut by the line

x = 2; the tangents at the points of intersection are concurrent at (l, µ). Then the value oflµ



E-26/28


3. If q is the angle between the line x 1 y 1 z 2

3 2 4

+ - -= = and the plane 2x + y – 3z + 4 = 0, then

cosec2q is equal to -4. Let ˆ ˆ ˆ ˆˆ ˆp i j, q j k, r k i= - = - = -

r r r . If sr is a unit vector such that [ ]p.s 0 q r s= =

r r r r r and

( )ˆ ˆ î j ks

+ l -m= ±

gr , then l +m + g

2ˆ ˆ î j k is

5. Let Sn be the area of the figure enclosed by a curve y = x2(1 – x)n (0 < x < 1) and x-axis,

if n

kn

k 1

1lim S

A®¥=

=å , then 'A' is equal to

6. The value of 150

13 49

01

3 50

0

(1 x ) dx

(1 x ) dx

æ ö-ç ÷

ç ÷ç ÷

-ç ÷ç ÷è ø

ò

òis

SAMPLE PAPER-1/Paper-2

E-27/28

ALLENSpace for Rough Work

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005

+91-744-2757575 [email protected] www.allen.ac.in

QUESTION PAPER FORMAT AND MARKING SCHEME :16. The question paper has three parts : Physics, Chemistry and Mathematics.17. Each part has two sections as detailed in the following table.

I have read all the instructionsand shall abide by them.

____________________________Signature of the Candidate

I have verified the identity, name and Formnumber of the candidate, and that questionpaper and ORS codes are the same.

____________________________Signature of the Invigilator

NAME OF THE CANDIDATE ................................................................................................

FORM NO. .............................................

Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-2

E-28/28

ALLEN

Que. No. Category-wise Marks for Each Question MaximumSection Type of Full Partial Zero Negative Marks of the

Que. Marks Marks Marks Marks section+3 0 –1

I(i) Single If only the bubble If none In allcorrect option 4 corresponding to — of the other 12

the correct option bubbles is casesis darkened darkened

+4 +1 0 –1One or more If only the bubble(s) For darkening a bubble If none In all

I(ii) correct 8 corresponding corresponding to each of the other 32option(s) to all the correct correct option, provided bubbles is casesoption(s) is(are) NO incorrect option darkened

darkened darkened+4 0 –1

Numerical If only the bubble If none In all II Value Type 6 corresponding — of the other 24

(Up to second to correct answer bubbles is cases decimal place) is darkened darkened

Your Target is to secure Good Rank in JEE 2021

Ïi;k bu funsZ'kks a dks /;ku ls i<+ s a

lkekU; %

1. ;g eksgjcU/k iqfLrdk vkidk iz'ui= gSA bldh eqgj rc rd u rksM+s tc rd bldk funsZ'k u fn;k tk;sA

2. iz'uksa dk mÙkj nsus ds fy, vyx ls nh x;h vkWfIVdy fjLikal 'khV (vks- vkj- ,l-) (ORS) dk mi;ksx djsaA

3. dPps dk;Z ds fy, bl iqfLrdk esa [kkyh LFkku fn;s x;s gSaA

4. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke o QkWeZ uEcj fyf[k, ,oa gLrk{kj cukb;sA

5. bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ysa fd blesa 28 i`"B gSa vkSj izR;sd fo"k; ds lHkh 18 iz'u vkSj muds mÙkj

fodYi Bhd ls i<+ s tk ldrs gSaA ;fn ugha] rks iz'ui= dks cnyus ds fy, fujh{kd ls lEidZ djsaA

6. ijh{kkFkhZ iz'ui= dks ijh{kk dh lekIrh ij ys tk ldrs gSaA

vkWfIVdy fjLikal 'khV (vks-vkj-,l-) %

7. vks- vkj- ,l- dks ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk,xkA

8. vks- vkj- ,l- esa gsj&Qsj@foÏfr u djsaA vks-vkj-,l- dk dPps dke ds fy, iz;ksx u djs aA

9. viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa esa dye ls fy[ksa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh fooj.k

vks-vkj-,l- es a dgha vkSj u fy[ksaA QkWeZ uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

vks-vkj-,l- ij cqycqyks a dks dkyk djus dh fof/k :

10. vks-vkj-,l- ds cqycqyksa dks dkys ckWy ikWbUV dye ls dkyk djsaA

11. cqycqys dks iw.k ± :i ls dkyk djsaA

12. cqycqys dks dkyk djus dk mi;qDr rjhdk gS :

13. vks-vkj-,l- e'khu tk¡P; gSA lqfuf'pr djsa dh cqycqys lgh fof/k ls dkys fd, x;sa gSaA

14. cqycqys dks rHkh dkyk djsa tc vki mÙkj ds ckjs esa fuf'pr gksA dkys fd, gq, cqycqys dks feVkus vFkok lkQ djus dk dksbZ

rjhdk ugha gSA

15. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksA

fujh

{kd

ds v

uqns'k

ksa d

s fcu

k eqgj

sa u r

ksM+s

Ïi;k 'ks"k funs Z'kks a ds fy, bl iqfLrdk ds vfUre i`"B dks i<+ s aA

Time : 3 Hours Maximum Marks : 204PAPER-2

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

SAMPLE PAPER-1

JEE(Advanced)

REVIEW TEST

Test Pattern

Hin

di

H-2/28

dPps dk;Z ds fy, LFkku

SOME USEFUL CONSTANTSAtomic No. : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s



H-3/28

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS


Hkkx-1 : Hkk Sfrd foKku[k.M-I(i) : (vf/kdre vad : 12)

� bl [k.M esa pkj iz'u gSa

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy , vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d iznhIr ySEi esa VaxLVu rUrq dk dk;Zdkjh rkieku 2450 K gS rFkk bldh mRltZdrk 0.30 gSA 25 w ds ,d ySEi ds

rUrq dk i`"Bh; {kS=Qy gS : (s = 5.67 × 10–8 S.I. bdkbZ)

(A) 0.1 cm2 (B) 0.2 cm2 (C) 0.3 cm2 (D) 0.4 cm2

2. nks yEcs rkjksa dks rkjksa ds yEcor~ fLFkj lekUrj iVfj;ksa ds ; qXe ij j[kk tkrk gSA iVfj;ksa ds e/; nwjh d, rkjksa ds e/; nwjh

x dh rqyuk esa vf/kd gSA rkj rFkk iVfj;k¡ izfr bdkbZ yEckbZ r izfrjks/k ds inkFkZ ls cuh gSA bl izdkj cus vk;r ds

yEcor~ ,d pqEcdh; ¶yDl ?kuRo B vkjksfir fd;k tkrk gS rFkk ,d rkj dks ,dleku pky v ls iVfj;ksa ds vuqfn'k

xfr djk;h tkrh gS tcfd nwljs dks fLFkj j[kk x;k gSA ;g ns[kk tkrk gS fd x ds ,d fof'k"V eku ij fLFkj rkj dks jksd

dj j[kus okyk vko';d cy lekIr gks tkrk gSA x Kkr dhft;sA

(A) 0v4m

pr (B) 0v2mpr (C) 02 vm

pr (D) 0v3mpr


H-4/28


3. iznf'kZr fp= esa CykWd ij fdruk {kSfrt cy F yxk;k tkuk pkfg;s rkfd ;g urry ij fu;r osx ls uhps dh vksj xfr dj

lds\ CykWd dk æO;eku m = 2 kg, CykWd o ry dh lrg ds e/; ?k"kZ.k xq.kkad 0.2 gS rFkk ry {kSfrt ds lkFk 45° dks.k

cukrk gSA

Fm

a

(A) 20 N (B) 13.33 N (C) 8 2N (D) 4 2N

4. ,d ,FkyhV nkSM+rk gqvk {kSfrt ds lkFk fdlh fo'ks"k dks.k ij ,d Hkkyk (Javelin) QSadrk gSA ;fn ,FkyhV ds lkis{k bl Hkkys

dh pky] iz{ksi.k ds le; ,FkyhV dh pky ds cjkcj gks rks ijkl vf/kdre izkIr djus ds fy;s ,FkyhV dks Hkkyk Lo;a ds lkis{k

fdl dks.k ij QSaduk pkfg;s\ ,FkyhV dh Å¡pkbZ dks ux.; ekuk tk ldrk gSA

(A) 45° (B) 30° (C) 60° (D) 37°


H-5/28


[k.M-I(ii) : (vf/kdre vad : 24)� bl [kaM esa vkB iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd

fodYi lgh gS(gSa)A� izR;sd iz'u ds fy,] iz'u dk (ds) mÙkj nsus gsrq lgh fodYi (fodYiksa) dks pqusA� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk %

iw.kZ vad : +4 ;fn dsoy (lkjs) lgh fodYi (fodYiksa) dks pquk x;k gSAvkaf'kd vad : +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSAvkaf'kd vad : +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gSa ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj

pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSaA'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ iz'u vuqÙkfjr gS)A½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyr gS_rks dsoy lHkh rhu lgh fodYiksa dk p;u djus ij gh +4 vad feys axs aA fcuk dksb Z xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,d dkspquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.k esa nwljkfodYi), –1 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

5. gkbMªkstu ln`'; ijek.kq ls ykbeu Js.kh ds laxr fofdj.k LiSDVªe ds ,d Hkkx dks ;gk¡ n'kkZ;k x;k gSA buesa ls ,d js[kk

vuqifLFkr gSA (R = 7 11610 m

15-´ )

1254 nm

25 nm

(A) ;g ijek.kq ghfy;e gSA (B) vuqifLFkr js[kk 3375

nm128

gSA

(C) iznf'kZr LiSDVªe] vo'kks"k.k LiSDVªe gSA (D) dqN ijek.kq] r`rh; mÙksftr voLFkk esa mÙksftr gksrs gSaA


H-6/28

6. nks L=ksr fp=kuqlkj ,d Jà[kyk eas tqM+s gq, gSA ,d [kqys ckg~; ifjiFk ds lkFk fcUnqvksa A o B ds e/; foHkokUrj V1 dh

x.kuk dhft;s] ;fn e1 = 5 V, r1 = 1 ohm, e2 = 3V, r2 = 2 ohms gksA ;fn dqath K dks can dj fn;k tk;s rks fcUnqvksa

A o B ds e/; foHkokUrj V2 D;k gksxk\ izfrjks/k R = 5 vkse gSA fLop dks can djus ds ckn cSVfj;ksa ls fuxZr /kkjk Hkh

Kkr dhft;sA

A Be1 1, r

e2 2, rK

R

(A) V1 = 4.3 V yxHkx (B) V2 = 3.8 V yxHkx

(C) cSVjh e1 ls fuxZr /kkjk yxHkx 1.2 A gSA (D) cSVjh e2 ls fuxZr /kkjk yxHkx 0.4 A gSA

7. izR;sd 1 kg æO;eku o 10 cm f=T;k okys nks le:i xksyksa dks 37° vkur dks.k okys urry ij leku Å¡pkbZ ls fojkekoLFkk

ls NksM+k tkrk gSA izFke xksys o f}rh; xksys ds fy;s ?k"kZ.k xq.kkad Øe'k% 0.1 o 0.3 gSA

37°

(A) vk/kkj rd igq¡pus ds fy;s izFke o f}rh; xksys }kjk yxk le; dk vuqikr 7591 gSA

(B) vk/kkj ij igq¡pus ij izFke rFkk f}rh; dh LFkkukUrj.kh; xfrt ÅtkZ dk vuqikr 1 : 1 gksxkA

(C) vk/kkj ij igq¡pus ij izFke rFkk f}rh; xksys dh pky dk vuqikr 9175 gSA

(D) vk/kkj ij igq¡pus ij izFke rFkk f}rh; xksyksa ds muds dsUæ ds lkis{k dks.kh; osx dk vuqikr 9175 gSA



H-7/28

8. ,d fLFkj yEch vkn'kZ ifjukfydk ds pkjksa bldh f=T;k ls vf/kd f=T;k R okyk ,d lek{kh; o`Ùkkdkj ywi fp=kuqlkj

?ksjs ds :i esa fo|eku gSA o`Ùkkdkj ywi ls /kkjk I izokfgr dh tkrh gSA fuEu esa ls fdl izdj.k esa ifjukfydk esa dksbZ fo|qr

okgd cy izsfjr ugha gksrk gS\

y

x

z

(A) ywi esa /kkjk fu;r j[kh tk;s rFkk ywi fLFkj gSA

(B) ywi esa /kkjk dk eku I = I0 sinwt gks rFkk ywi fLFkj gksA

(C) ywi esa /kkjk fu;r gks rFkk ywi blds O;kl (y-v{k) ds lkis{k dks.kh; ljy vkorZ xfr djrk gS (ijUrq ifjukfydk dh lrg

rd ugha igq¡prk gS)(D) ywi esa /kkjk fu;r gks rFkk ywi bldh v{k ds yEocr~ fn'kk esa bldh ek/; fLFkfr ds lkis{k ljy vkorZ xfr djrk gS

(ijUrq ifjukfydk dks ØkWl ugha djrk gSA)

9. æO;eku 5 kg okyh ,d Å/okZ/kj yVdrh gqbZ NM+ A ij ,d dhyd ls yVdh gqbZ gSA NM + ds e/; fcUnq ls ,d

æO;ekughu vforkU; jLlh tqM+h gqbZ gSA bls fp=kuqlkj Å/okZ/kj ls 37 fMxzh dks.k ij [khapk tkrk gSA nwljs fljs ij 20 kgdk ,d æO;eku yVd jgk gSA fudk; dks lkE;koLFkk esa cuk;s j[kus ds fy;s vkjksfir cy F1 o F2 dk eku rFkk jLlh esa

ruko T o dhyd ij cy F Kkr dhft;sA

20kgF1

F2

A

37°

(A) F1 = 75 N (B)F2 = 200 N (C) T = 250 N (D) F = 25 10N



H-8/28

10. yEckbZ L o æO;eku m okyh ,d NM+ dks leku yEckbZ L okyh nks jfLl;ksa dh lgk;rk ls Nr ls yVdk;k tkrk gSA ;g

nks izdkj ls nksyu dj ldrh gS% (1) dkxt ds ry ds yEcor~ rFkk (2) dkxt ds ry esaA ;s jfLl;k¡ æO;ekughu o

vforkU; gSA izR;sd izdj.k esa vkorZdky Kkr dhft;sA

LL

L

(A) izdj.k (1) esa = p3 L

T 22 g

(B) izdj.k (2) esa = p3 L

T 22 g

(C) izdj.k (1) esa 5LT 2

3 3g= p (D) izdj.k (2) esa 5L

T 23 3g

= p

11. fdlh h eksVkbZ dh ,d vuUr leryh; ijr esa ls h O;kl dh ,d xksyh; xqfgdk dkV dj fudky yh tkrh gSA ;g ijr

vk;ru vkos'k ?kuRo r ls ,dleku :i ls vkosf'kr gSA

A

B

(A) A ij fo|qr {ks= 0

h

6

re nka;h vksj gSA

(B) B ij fo|qr {ks= 0

h

3

re Åij dh fn'kk esa gSA

(C) ;fn xqfgdk ds dsUæ ij foHko 'kwU; gks rks A ij foHko 2

0

h12r

e gksxkA

(D) ;fn xqfgdk ds dsUæ ij foHko 'kwU; gks rks B ij foHko 2

0

13 h12

re gksxkA



H-9/28

12. Hkqtkvksa 4m × 3m × 1m okyk ,d f=Hkqtkdkj fizTe ty ls iw.kZr;k Hkjk gqvk gSA ok;qe.Myh; nkc dks ux.; ekuus ij

ty (?kuRo = 1000 kg/m3) }kjk yxk;k x;k nkc

A

B

D

E

F

C3m

4m

1m

(A) Qyd ABCD ij 'kwU; gSA

(B) Qyd CDEF ij 2 × 104 N gSA

(C) Qyd ADE ij 8 × 104 N gSA

(D) Qyd ABEF ij 10 × 104 N gSA



H-10/28

[kaM-II : (vf/kdre vad : 24)� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_ mnkgj.k

6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lgh gksxsa)dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksa dks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA



1. ,d [kks[kys xksys dh ckg~; o vkUrfjd f=T;k (4.20 ± 0.01) cm o (3.90 ± 0.01) cm ekih tkrh gSA vk;ru ds ekiu

esa vf/kdre izfr'kr =qfV gksxhA

2. R f=T;k o m nzO;eku dh ,d oy; Q vkos'k ls le:i vkosf'kr gSA ;g n'kkZ;s vuqlkj 'kq¼ :i ls yq<+d jgh gSA B ifjek.k

dk ,d le:i pqEcdh; {kS= oy; ds ry ls yEcor~ mifLFkr gSA ;fn /kjkry }kjk vkjksfir vfHkyEc cy lmg gS, rks l

Kkr dhft;sA QvB = mg yhft;sA

+ + ++

++

+++

+++

v



H-11/28

3. uhps fn;s x;s ifjiFk ij fopkj djsaA izkjEHk esa fLop S, le; 0.2 ln 2 s ds fy;s fLFkfr 1 esa gSA fQj fLop dks ifjofrZr

djds fLFkfr 2 ij dj fn;k tkrk gSA le; t (fLop ds fLFkfr ifjorZu ls ekik x;k) ds i'pkr~ 5 kW izfrjks/k ds fljksa ij

oksYVrk 35

oksYV ik;h tkrh gS] rks t dk eku (feyh lSd.M esa) gS

10V 2

1S

5kW

10kW

20 Fm

4. uhps n'kkZ;s x;s fp= esa] og vf/kdre lEHko vKkr izfrjks/k (X) tks iksLV vkWfQl ckWDl }kjk ekik tk ldrk gS] Xmax gS

rFkk ftldk eku R × 105 W gS] rks R gS : (bl iz;ksx esa] ge Hkqtk AB esa dsoy ,d Iyx fudkyrs gSa o Hkqtk BC esa ge

dsoy ,d Iyx fudkyrs gS ysfdu Hkqtk AD esa ge dbZ Iyx fudkyrs gSa) :

G+–

E

1000 100 10 B10 100

D

A

500200200100 X

unknownresistance

QP

R

C1000

5000 2000 1000



H-12/28

5. O;kl 2 mm o 1mm okyh nks vkil esa tqM+h gqbZ nks ds'kufy;ksa ds æoksa ds Lrjksa esa vUrj (mm esa) Kkr dhft;sA æo dk

i`"B ruko 0.07 N/m gSA nksuksa ds'kufy;ksa ds fy;s uopUæd laidZ dks.k 'kwU; gSA æo ?kuRo 1000 kg/m3 gSA

6. izksVkWu i`FkDdj.k ÅtkZ] ÅtkZ dh og ek=k gS tks ge ,d ukfHkd ls ,d izksVkWu dks gVkus ds fy;s iznku djrs gSA He42 dh

izksVkWu i`FkDdj.k ÅtkZ (MeV esa) Kkr dhft;sA

( )Hm 11 dk nzO;eku = 1.007825 amu

( )Hm 31 dk nzO;eku = 3.016049 amu

( )Hem 42 dk nzO;eku = 4.002603 amu

(1amu = 931 MeV ysa)dPps dk;Z ds fy, LFkku


H-13/28

Hkkx-2 : jlk;u foKku[k.M-I(i) : (vf/kdre vad : 12)


� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA





1. ,d ik= esa 20 atm ij He xSl rFkk 'P' atm ij CH4 xSl mifLFkr gSA tc ik= esa ,d lw{e fNæ fd;k tkrk gS rks izkjEHk

esa ckgj vkus okyh xSlksa esa eksy }kjk 80% He xSl mifLFkr gS rks 'P' dk eku gS(A) 1 atm (B) 40 atm (C) 2.5 atm (D) 10 atm

2.

OH

O

dil.NaOH¾¾¾¾¾®

vfHkfØ;k feJ.k esa mifLFkr eq[; :i gS

(A)

OH

O(B)

O

O

(C)

O

O(D)

O

C – ONa

O



H-14/28

3. fuEu esa ls dkSulh vfHkfØ;k] vfUre mRikn ds :i esa dkcks ZfDlfyd vEy nsrh gS

(A) O

O

O 3(i)H O(ii)Δ

Å¾¾¾¾® (B) CH – CH – C – H3 2

O mcpba

Δ¾¾¾¾®

(C) O O

OEt3(i)H O

(ii)Δ

Å¾¾¾¾® (D)

32

–

(i) O(ii) Zn,H O(iii)O H,Δ

¾¾¾¾¾®

4. nh x;h vfHkfØ;k esa fuEu esa ls dkSu] mRikfnr ugha gksrk gS?

lksfM;e veyxe water¾¾¾¾®

(A) H2 (B) NaOH (C) Hg (D) O2



H-15/28






� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi ), –1 vad feysaxs] pkgs lgh fodYi (fodYiksa ) dks pquk x;k gks ;k u pquk x;k gksA

5. lgh dFkuks dk p;u dhft;s

(A) tc foy;u dks ruq fd;k tkrk gS rks lfØf;r pkjdkWy ij vf/kd foys; vf/k'kksf"kr gksrs gS(B) lHkh dksyksbMh foy;u izHkkoh :i ls fV.My izHkko iznf'kZr djrs gS

(C) viksgu (Dialysis) dk iz;ksx lkWy ds LdUnu ds fy;s fd;k tk ldrk gS

(D) nzoLusgh lkWy dh rqyuk esa nzofojks/kh lkWyksa dks vklkuh ls LdfUnu fd;k tk ldrk gS



H-16/28

6. oksYVehVj esa 30 lSd.M esa fu{ksfir /kkrq dk æO;eku 200 gms gSA uhps fn[kk;s x;s /kkjk v/s le; dk vkjs[k dkfo'ys"k.k dhft;s rFkk lgh dFkuksa dk p;u dhft;s :

i

100 mA

10sec 20sec 30sec

t

(A) /kkrq dk oS|qrjklk;fud rqY;kad 100 g/dwykWe gSA(B) igys 10 lSd.M esa 33.33 xzke fu{ksfir gq;h(C) 66.66 mA dh fu;r /kkjk ls leku le; es yxHkx leku ek=k es fu{ksfir gksxhA(D) igys 15 lSd.M es 100 gms /kkrq fu{ksfir gq;hA

7. 4 atm rFkk 27°C ij vfHkfØ;k : A(s) �� 2B(g) ] lkE; esa gSA fu;r rkiØe ij ;fn ra= dk vk;ru c<+rk gS rksu;s lkE; ij] izkjfEHkd lkE; dh rqyuk esa :-(A) A(s) ds eksy ?kVrs gS (B) B(g) ds eksy c<+rs gS(C) A(s) dh eksyj lkUnzrk ?kVrh gS (D) B(g) dh eksyj lkUnzrk ?kVrh gS

8. fuEu eas ls fdls eqDr ewyd izfrLFkkiu vfHkfØ;k esa oxhZÏr (classified) fd;k x;k gS

(A) R – C – OAgO

2 4Br ,CClΔ¾¾¾¾¾® (B)

2 2

HBrR O ,hν¾¾¾¾¾®

(C) NBSΔ¾¾¾® (D) 2

4

BrCCl¾¾¾®



H-17/28

9. ,sfYMgkbM ds fuekZ.k ds fy, fuEu esa ls fdldk iz;ksx fd;k tkrk gS?(A) LVsQsu vip;u dk

(B) jkstsueq.M vip;u dk

(C) DyhesU'ku vip;u dk

(D) Xysf'k;y , sflfVd vEy esa Øksfed , sugkbMªkbM }kjk 1-izksisukWy dh vfHkfØ;k dk10. fuEu esa lgh Øe pqfu, :-

(A) F > N > C > B > Si (v/kkfRod y{k.k)(B) F2 > Cl2 > O2 > N2 (vkWDlhdkjd xq.k ds in esa jklk;fud lfØ;rk)(C) K > Mg > Al > B (/kkfRod y{k.k)(D) Ne > F > O > N (izFke vk;uu ÅtkZ)

11. fuEu eas ls dkSu ls fodYi esa] nh x;h Lih'kht ds jax ds fy, lgh Li"Vhdj.k iznf'kZr fd;k x;k gS?(A) FeSO4.7H2O Þ t2g ls eg laØe.k(B) Br2 = s* ls p* laØe.k(C) KMnO4 = L ls M laØe.k(D) gkbMªkstu ijek.kq dk ckej Js.kh mRltZu Þ n = 3,4...... ls n = 2 esa bysDVªkWu dk vuqÙkstu

12. fuEu esa ls dkSu] ty esa vfoys;'khy gS@gSa?(A) fo"keyEck{k lYQj (B) gkbMªkstu czksekbM(C) csjk;Vh yo.k (D) Qsfjd DyksjkbM



H-18/28

[kaM-II : (vf/kdre vad : 24)� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_

mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lghgksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA


0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••




1. XywVkfFkvksu (vkf.od Hkkj = 307 gm/mol) ,d egRoiw.kZ izfrvkWDlhdkjd gS tks eqDr ewydksa }kjk dksf'kdh; ?kVdksadks gksus okyh {kfr dks jksdrk gS tgk¡ ;g fuEu izdkj f}ydhÏr gksrk gS

HS

NH

NH2

HO

O O

NH

HO

O

O

¾¾® OHN

HOO

SS

NHNH2

HO

O O

NH

HO

OO

HNH2N

OH

OO

XywVkfFkvksu dk gm rqY;kad Hkkj dh x.kuk dhft,

2. 1 M HA (pKa = 4.2) rFkk 1 M NaA dks fefJr djds ,d foy;u cuk;k x;kA ;fn foy;u dh pH = 4.5 gaSA

NaA rFkk HA ds feyk;s x;s vk;ruksa dk vuqikr crkb;saA (fn;k gS : log 2 = 0.3).dPps dk;Z ds fy, LFkku


H-19/28

3. fdrus ;kSfxd NaNO2 + HCl ds lkFk ihyk æo nsrs gSa

(i) CH3 – CH2 – NH2 (ii) N – CH3

H(iii) NH2 (iv) N

H

(v) CH – N – CH3 3

CH3(vi) CH – – N – CH3 3CH2

H

(vii) CH – – NH – CH3 3C

O

(viii) CH – – NH3 2C

O

4. fuEu eas ls dqy la'ysf"kr la?kuu cgqyd fdrus gSa :(i) XyhIVy (ii) PVC (iii) lsyqykst (iv) MsØksu

(v) vkWjykWu (vi) ukbykWu-66 (vii) csdsykbV

(viii) ;wfj;k QkesZfYMgkbM jsflu5. 3d Js.kh dh /kkrqvksa dh lokZf/kd lkekU; vkWDlhdj.k voLFkk] +2 gS ysfdu Sc+2 , okLro esa vKkr gS 3d Js.kh ds

lHkh Kkr M+2 vk;uksa esa mifLFkr 3d-bysDVªksuksa dk ;ksx crkbZ;s6. ,d fdjsy ladqy ;kSfxd [Co(Co(NH3)4(OH)2)3]Br6 dh f=foeh; lajpuk esa Co–N rFkk Co–O ca/kuksa dh dqy

la[;k crkbZ;sdPps dk;Z ds fy, LFkku


H-20/28


Hkkx-3 : xf.kr[k.M-I(i) : (vf/kdre vad : 12)


� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy , vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA





1. O;atd y = 10sin2x + 6sinx cosx + 14 cos2x ds vf/kdre rFkk U;wure ekuksa dk ;ksxQy gksxk -(A) 12 (B) 24 (C) 36 (D) 48

2. vUrjky x Î [0,2p] esa vlfedk 2sin2x – 5sinx + 2 > 0 ds iw.kk±d gyksa dh la[;k gksxh-(A) 3 (B) 4 (C) 5 (D) 6

3. ml js[kk dh izo.krk dk fujis{k eku] tks nksuksa oØksa y = x2 rFkk x2 + y + 1 = 0 dh mHk;fu"B Li'kZ js[kk gS] gksxk&

(A) 5 (B) 2 (C) 3 (D) 2

4. lfEeJ la[;k;sa z1 rFkk z2 bl izdkj gS fd |z1| = 3, |z2| = 5 rFkk |z1 + z2| = 7 gSA rc 6162

zarg

z dk eku fuEu gks

ldrk gS -

(A) p (B) 2

3

p(C)

3

p(D) 2p


H-21/28







� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –1 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

5. ekuk k Î Q rFkk ƒ(x) = x2 – (3 – 2k)x + 2 – 4k gks] rks lgh dFkuksa dk p;u dhft, -(A) ƒ(x) = 0 ds ewy ifjes; gSA

(B) ;fn ƒ(x) = 0 ds ewy foijhr fp ds gks] rks 1

k2

> gksxkA

(C) ;fn 1k

2£ gks] rks lehdj.k ƒ(x) = 0 ds ewy lnSo leku fp ds gksaxsA

(D) ;fn ƒ(x) = 0 dk Bhd ,d ewy 2 ls cM+k gks] rks k dk vf/kdre iw.kk±d eku –1 gksxkA


H-22/28


6. fp=kuqlkj lery esa 7 fcUnq fn;s x;s gS] rc mu 7 fcUnqvksa ds iz;ksx ls&1

2

34 5 6 7

(A) cuk;h tk ldus okyh js[kkvksa dh la[;k 10 gksxhA(B) cuk;s tk ldus okys f=Hkqtksa dh la[;k 24 gksxhA(C) cuk;s tk ldus okys prZqHkqtksa dh la[;k 6 gksxhA(D) cuk;s tk ldus okys o`Ùkksa dh vf/kdre la[;k 24 gksxh] tgk¡ izR;sd oÙk ij 7 esa ls de ls de rhu fcUnq fLFkr gksA

7. 3 FkSys A, B rFkk C gSA FkSys A esa 1 yky ,oa 2 gjh xsnsa] FkSys B esa 2 yky ,oa 1 gjh xsan rFkk FkSys C esa dsoy 1 gjh xsangSA FkSys A ls 1 xsan fudky dj mldks FkSys B esa j[krs gS] mlh le; FkSys B ls 1 xsan fudky dj FkSys C esa j[krs gS rFkkvUr esa FkSys C ls 1 xsan fudky dj FkSys A esa j[krs gSA ;g izfØ;k iw.k± gksus ij izkf;drk rkfd FkSys A esa 2 yky rFkk 1gjh xsan gks] gksxh -

(A) 1

4(B) 1

2(C)

1

3(D)

1

68. ;fn A rFkk B dksfV 3 ds oxZ vkO;wg bl izdkj gS fd AB = aA + bB, tgk¡ a, b Î R – {0} gks] rks fuEu esa ls

dkSulk@dkSuls lgh dFku gksxk@gksaxs(A) A rFkk B Øefofues; gSA (B) A – bI O;qRØe.kh; gSA(C) B – aI O;qRØe.kh; gSA (D) A = B


H-23/28

9. lHkh /kukRed la[;kvksa x,y ds fy;s Qyu ƒ(x), laca/k ƒ(xy) = ƒ(x).ƒ(y) dks larq"V djrk gS rFkk ƒ(2) = 10 gS] rks(ƒ–1(x), ƒ(x) ds izfrykse dks n'kkZrk gS) -

(A) ( )4

r

r 0

ƒ 2 10000=

=å (B) ( )4

r

r 0

ƒ 2 11111=

=å

(C) ( )4

1 r

r 0

ƒ 10 31-

=

=å (D) ( )4

1 r

r 0

ƒ 10 32-

=

=å

10. ;fn 2

1x 2 21 cosx

x 0lim(5 x ) L.e-

®+ = gks] rks

1t t t t t 2t

tlim(1 2 3 ......... 25 L )®¥

+ + + + + fuEu ls vf/kd gksxk

(A) 1 (B) 3 (C) 5 (D) 30



H-24/28

11. ekuk vUrjky [a, b] esa larr~ Qyu ƒ(x) bl izdkj gS fd ƒ(a) ¹ ƒ(b) rFkk ƒ(a), ƒ(b) > 0 gks] rks fuEu esa ls dkSulk dFkulgh gksxk \

(A) ;fn ƒ(c) =ƒ(a) ƒ(b)

2

+ gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(B) ;fn ƒ(c) ƒ(a)ƒ(b)= gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(C) ;fn 5ƒ(c) 3ƒ(a) 2ƒ(b)= + gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(D) ;fn 2ƒ(c) 3ƒ(a) ƒ(b)= - gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

12. ;fn ( )3xP e cos 2013x dx= ò , ( )3xQ e sin 2013x dx= ò (tgk¡ P rFkk Q nksuksa dk lekdyu vpj 'kwU; gS) rFkk

ƒ(x) = ln((P2 + Q2) (32 + 20132)) gks] rks(A) ƒ(1) + ƒ(2) = 18 (B) ƒ'(1) = ƒ'(–1)(C) ƒ(x) fo"ke ?kkr cgqin gksxk (D) ƒ(x), x = 0 ds lkis{k lefer gksxk



H-25/28

[kaM-II : (vf/kdre vad : 24)� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_

mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lghgksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA


0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••




1. ;fn P1, P2, P3 nh?kZoÙk 3x2 + y2 – 12 = 0 ij fLFkr fcUnq gS rFkk P1', P2', P3' lgk;d o`Ùk ij fLFkr buds laxr

fcUnq gS] rks f=Hkqt P1' P2' P3' dk {ks=Qy] f=Hkqt P1P2P3 ds {ks=Qy dk l xquk gks] rks l2 dk eku gksxk

2. ;fn vody lehdj.k 3dyx y x

dx+ = ds iw.kk±d oØksa dk fudk;] js[kk x = 2 }kjk izfrPNsfnr gS] izfrPNsn fcUnqvkas ij

[khaph xbZ Li'kZ js[kk;sa] fcUnq (l, µ) ij laxkeh gks] rks lµ dk eku gksxk



H-26/28


3. ;fn js[kk x 1 y 1 z 2

3 2 4

+ - -= = rFkk lery 2x + y – 3z + 4 = 0 ds e/; dks.k q gks] rks cosec2q dk eku gksxk

4. ekuk ˆ ˆ ˆ ˆˆ ˆp i j, q j k, r k i= - = - = -r r r gSA ;fn s

r bdkbZ lfn'k bl izdkj fd [ ]p.s 0 q r s= =r r r r r

rFkk ( )ˆ ˆ î j k

s+ l - m

= ±g

r

gks] rks l + m + g2ˆ ˆ î j k gksxk

5. ekuk Sn, oØ y = x2(1 – x)n (0 < x < 1) rFkk x-v{k }kjk ifjc¼ vkjs[k dk {ks=Qy gS] ;fn n

kn

k 1

1lim S

A®¥=

=ågks] rks 'A' dk eku gksxk

6. 150

13 49

01

3 50

0

(1 x ) dx

(1 x ) dx

æ ö-ç ÷

ç ÷ç ÷

-ç ÷ç ÷è ø

ò

ò dk eku gksxk

H-27/28

SAMPLE PAPER-1/Paper-2ALLENdPps dk;Z ds fy, LFkku

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005

+91-744-2757575 [email protected] www.allen.ac.in

iz'ui= dk izk:i vkSj vadu ;kstuk %16. bl iz'ui= esa rhu Hkkx gSa % HkkSfrd foKku] jlk;u foKku ,oa xf.krA17. izR;sd Hkkx esa nks [k.M gSa ftudk fooj.k fuEufyf[kr rkfydk esa fn;k x;k gSA

eSaus lHkh funsZ'kks a dks i<+ fy;k gS vkSj eS mudk

vo'; ikyu d:¡xk@d:¡xhA

____________________________

ijh{kkFkhZ ds gLrk{kj

eSaus ijh{kkFkhZ dk ifjp;] uke vkSj QkWeZ uEcj dks iwjh rjgtk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks aleku gSaA

____________________________

fujh{kd ds gLrk{kj

ijh{kkFkh Z dk uke ................................................................................................

QkWeZ uEcj .............................................

H-28/28

Target : JEE (Main + Advanced) 2020/13-09-2020/Paper-2ALLEN

[k.M iz'u dk iz'uks a oxkZuqlkj izR;sd iz'u ds vad [k.M es aizdkj dh iw.kZ vad vkaf'kd vad 'kwU; vad ½.k vad vf/kdÙke

la[;k vad I(i) ,dy lgh +3 0 –1

fodYi 4 ;fn flQZ lgh fodYi ds — ;fn fdlh Hkh vU; lHkh 12vuq:Ik cqycqys dks cqycqys dks dkyk ifjfLFkfr;ksa

dkyk fd;k gS ugha fd;k gS esa,d ;k ,d ls +4 +1 0 –1

I(ii) vf/kd lgh ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i ;fn fdlh Hkh vU; lHkhfodYi 8 (fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] cqycqys dks ifjfLFkfr;ksa 32

cqycqys (cqycqyksa) dks ;fn dksbZ xyr fodYi dkyk dkyk ugha esa dkyk fd;k x;k gS ugha fd;k gS fd;k gS

la[;kRed eku +4 0 –1 II izdkj ;fn flQZ lgh mÙkj ds ;fn fdlh Hkh vU; lHkh

(n'keyo ds nks 6 vuq:Ik cqycqys dks — cqycqys dks ifjfLFkfr;ksa 24

LFkku rd) dkyk fd;k gS dkyk ugha esafd;k gS

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Tes Pttern Eglish REVIEW TEST CLAS ONT PROGRAMME ... - ALLEN