PAPER-1 - Doubtion

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

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Time : 3 Hours Maximum Marks : 198

)1001CJA545120002)(1001CJA545120002) Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

JEE(Main+Advanced) : LEADER LIVE PHASE-(I + II) & LEADER (ON LINE + OFF LINE) PHASE-(I + II)

JEE(Advanced)REVIEW TEST

08-11-2020

PAPER-1

En

glis

h

E-2/28

ALLEN

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

1cm

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

1001CJA545120002

Target : JEE (Main + Advanced) 2021/08-11-2020/Paper-1

Leader Live Phase-(I + II) & Leader (On Line + Off Line) Phase-(I + II)/08-11-2020/Paper-1ALLEN

E-3/281001CJA545120002

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

PART-1 : PHYSICSSECTION–I(i) : (Maximum Marks : 18)

� This section contains SIX 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. A child is sitting in a chair connected to a rope that passes over a frictionless pulley. Thechild pulls on the loose end of the rope with a force of 250 N. The child’s weight is 320 N andthe chair weighs 160 N. The child is accelerating. The force that the seat of the chair exertson the child is approximately

(A) 83 N (B) 70 N (C) 13 N (D) 57 N


Target : JEE (Main + Advanced) 2021/08-11-2020/Paper-1ALLEN

E-4/28 1001CJA545120002


2. In the diagram, BAC is a rigid fixed rough wire and angle BAC is 60º. P and Q are two identicalrings of mass m connected by a light elastic string of natural length 2a and force constant

mga . P and Q are in equilibrium when PA = AQ = 3a. The least coefficient of friction between

the ring and the wire is m. The value of m + 3 is

3a

CB

30°30°3a

A

QP

(A) 2 (B) 3 (C) 4 (D) 73. A man can swim at 2 m/s in a swimming pool. He enters a 100 m wide river at one bank and

swims at right angles to the river flow velocity (w.r.t. water). If his drift on reaching theother bank is 40 m, calculate the river flow velocity.(A) 0.4 m/s (B) 0.8 m/s (C) 0.6 m/s (D) 2 m/s


E-5/281001CJA545120002


4. Reading shown in two spring balances, S1 and S2 is 60 kg and 30 kg respectively. Spring S1 isin elongated state & spring S2 is in compressed state. Lift is accelerating upward withacceleration 10 m/s2. The mass is stationary with respect to lift. Then the mass of the blockwill be :-

m

S1

S2

10 m/s2

(A) 45 kg (B) 30 kg (C) 60 kg (D) 15 kg5. A particle performs circular motion of radius 1m from rest. The tangential accelration of the

particle at any time t is given by at = t ms–2. The radial acceleration of the particle at t = 2 secis :-(A) 1 ms–2 (B) 2 ms–2 (C) 0.5 ms–2 (D) 4 ms–2

6. A projectile is thrown with a speed u at angle q with the horizontal. The radius of the curvature

when the velocity makes an angle 2qæ ö

ç ÷è ø

with the horizontal is :

(A)2 2

3u cos

2g cos ( / 2)q

q (B) 2 2

3u sin

g sin ( / 2)q

q (C) 2 2

3u cos

g cos ( / 2)q

q (D) 2 2

32u sin

g sin ( / 2)q

q


E-6/28 1001CJA545120002

SECTION-I(ii) : (Maximum Marks: 24)� This section contains SIX 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 : –2 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 correctoption(s) will result in –2 marks.

7. Which of the following is accelerating?(A) A pendulum swinging back and forth.(B) A runner maintaining a constant speed about an circular track.(C) A ball rolling down an incline.(D) A toy car travelling at its maximum speed in a straight line.



E-7/281001CJA545120002

8. A projectile strikes the inclined plane perpendicularly as shown in the figure. Which of thefollowing is/are CORRECT?

aO

(A) Magnitude of position vector joining point O to the projectile never decreases(B) Time taken to reach the highest point above horizontal is more than half the time of flight(C) During projectile motion instantaneous velocity can never be perpendicular to initial

velocity vector(D) After collision with inclined plane projectile can never move in vertical direction

9. Two particles are projected from position ‘A’ and ‘B’ seperated by a distance of 70 m onhorizontal ground. If both the particles collides in mid air. Select the correct statement [Bothparticles are projected simultaneously]

A B45°

50m/s 40 2 m/sÖ

q

70m

(A) They will collide if angle of projection for ‘A’ is 37°(B) Particles will not collide for any value of q.(C) Particles will collide at t = 1 sec after projection(D) They will collide if angle of projection for ‘A’ is 53°



E-8/28 1001CJA545120002

10. In the arrangement shown, all surfaces are frictionless. AB part of the string is horizontaland BC part is parallel to the incline plane. System is released from rest. During the motionof block along incline it is found that normal contact force between m and M is zero, accelerationof wedge is a and tension in the string is T, then which of the following is/are CORRECT :

mC

BA

Mq

(A) T = (m + M) a (B) ( )MaT

1 cos=

- q(C) a = g cot q (D) T = ma + mg sin q

11. A car is moving without slipping anticlockwise sense (top view of track) in a circular path ona rough horizontal surface. rv and ra are velocity and acceleration vector at time 't'. When caris at point P, the direction of friction force for different conditions mentioned in the optionsbelow are given by arrows as shown in figure. Select the CORRECT alternative(s).

r

5

12

4

3

v®

P

Top view of the circular track

(A) If × =r ra v 0 , friction force will act along arrow 3.

(B) If × >r ra v 0 , friction force will act along arrow 2.

(C) If × <r ra v 0 , friction force will act along arrow 5.

(D) If × <r ra v 0 friction force will act along arrow 4.



E-9/281001CJA545120002

12. A block A moves with a velocity v on a plank B which is moving with velocity 2v. If the groundis smooth and friction is present between A and B. (Here Wf = Total work done by friction onblock A & block B ; (Wf)A , (Wf)B are work done by friction on block A & B respectively)

AB

v

2v

(A) (Wf)A is positive(B) (Wf)B is negative(C) Wf = WfA + WfB is negative(D) Wf does not depend on the reference frame



E-10/28 1001CJA545120002

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 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

••••••••••

� 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 In all other cases.

1. A particle moves in rectilinear motion. Given the a-s curve shown in figure. Initial velocity ofthe particle is 10m/s. Velocity of the particle after the particle has travelled 30m is X m/s.The value of X/10 is

15 30

10

a(m/s )2

s(m)



E-11/281001CJA545120002

2. A boat starts from A with velocity brVr

making an angle a = 30º with line AB and crosses the

river. Velocity of river flow is rVr

as shown in figure. The line AC makes an angle b = 60º with

line AB. If the boat always remains on line AC then, the value of br

r

VV

is

Vr

CB

A

Vbr

3. The system is released from rest when spring is at its natural length. Spring constant is100N/m and mass of each block is 10kg. The velocity 'v' of block A as a function of position x isgiven as v2 = ax – bx2. The value of a/b is equal to (g = 10 m/s2)



E-12/28 1001CJA545120002

4. A force of constant magnitude F = 10 N acts on a particle moving in a plane such that it is

perpendicular to the velocity ( )v v v 5m / s= =r r of the body, and the force is always directed

towards a fixed point. Then the angle (in radian) turned by the velocity vector of the particleas it covers a distance S = 10m is : (Take : mass of the particle as m = 2kg)

5. A particle moves according to equation x = 10 sin2pt, y = 10 cos 2pt. Find the work performedby the net force from t = 0 to t = 5 seconds.



E-13/281001CJA545120002

6. The block has mass M and rests on a surface for which the coefficients of static and kineticfriction are ms and mk respectively. A force F = kt2 is applied to the cable. Velocity of block att = 2 sec is v m/s. Then value of 23/v is. Given : M = 24 kg, k = 60 N/s2, ms = 0.5, mk = 0.4

FM



E-14/28 1001CJA545120002

PART-2 : CHEMISTRYSECTION–I(i) : (Maximum Marks : 18)




1. Correct order for acidic strength :

(A)NO2

COOH>

Cl

COOH>

CN

COOH

(B) Cl3C—COOH > CH3COOH > Br3C—COOH(C) CH3—COOH < CH2 = CH—COOH < HCºC—COOH(D) All of these

2. Correct order for carbanion stability :

(A) > >

1 1 1

(B) H CCH3

CH3

< CH3 CCH3

CH3

1 1

(C) CH2 < CH2

1 1(D) >

1 1

3. What is the INCORRECT bond angle order for the following molecule?(A) PF3 > PH3 (B) NH3 > NF3

(C) CH3 CH3

O OCH3 Ha b> (D) 2

3CO - > CO2 > 4NH+



E-15/281001CJA545120002

4. Which of the following species is different by their geometry and shape?(A) XeO3F2 (B) IF7 (C) 2

6SiF - (D) 5XeF1

5. A liquid solution having two liquids A & B. If vapour pressure of solution is 400 mm Hg. GivenPA

0 = 600 mmHg ; PB0 = 300 mmHg, then mol fraction of A in distillate is

(A) 1/2 (B) 1/3 (C) 2/3 (D) 1/4

6. An element has a fcc structure with the cell edge 'a' pm. Density of element is r gm/cc.How many unit cell are present in W gm of the element ? [NA = Avogadro's No.]

(A) 30

3W 10a 4

´´ r ´ (B) 30

3W 10

a´

´ r (C) 30

3A

W 10N( 2 a)

´r (D)

36

3A

W 10a N

+´r



E-16/28 1001CJA545120002








7. Choose the incorrect pairs from the following -(A) Monoclinic | a ¹ b ¹ c | face centre(B) Rhombohedral | a = b = c | primitive(C) Hexagonal | a = b = 90, g = 120º | body centred(D) Orthorhombic | a = b ¹ c | primitive

8. For the reaction A + 3B ® C, select the correct statement -

(A) d[C] d[A]dt dt

-=

(B) 3d[C] d[B]dt dt

= -

(C) Rate law must be r = k [A][B]3

(D) Units for rate of reaction are independent of order of reaction


E-17/281001CJA545120002

9. Which of the following statements are incorrect?(A) HCN has s & p bond ratio is 1 : 1(B) Bond energy order Cl2 > F2 > Br2 > I2

(C) PbF4 & BiF5 are do not exist(D) NH3 is better Lewis Base than PH3

10. Which of the following statements are correct?(A) PCl5(s) ® cationic part (sp3 hybridised) + anionic part (sp3d2 hybridised)(B) PBr5(s) ® cationic part (sp3 hybridised) + anionic part (noble gas configuration)(C) I2Cl6(l) ® cationic part (sp3 hybridised) + anionic part (sp3d2 hybridised)(D) N2O5(s) ® cationic part (linear) + anionic part (trigonal planar)

11. Identify correct stability ?

(A) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

(B) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

– – – –

(C) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

+ + + +

(D) None of theseSpace for Rough Work


E-18/28 1001CJA545120002

12. Identify correct order ?

(A) H—F < H—Cl < H—Br < H—I Acidic strength

(B) CH4 < NH3 < H2O < H—F Acidic strength

(C)

NH2

<

NH2

Basic strength

(D)

Cl

F

OBr–

>F

Cl

O–

Br

>Cl

Br

O–

FStability order



E-19/281001CJA545120002



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. How many p-electrons are involve in conjugation -

2. How many resonating structures are possible for OH



E-20/28 1001CJA545120002

3. Number of molecules having both pp-pp & pp-dp bond.COCl2, SO2, SO3, B3N3H6, XeO4, POCl3

4. Find the number of chemcial species which does not exist.NCl5, SI6, TlI3, FeI3, CuI2, PbI4, BiI5, NOF3

5. Equal mass of A & B are present in liquid solution then total pressure exerted by vapoursis X torr , Find (X/6). [A = 80 gm/mol, B = 120 gm/mol]

Pr(torr)

30

0100

1000

PT

PB PA

Mass % of A

40

Pr(torr)

Mass % of B

6. How many number of 3rd nearest neighbour atom in a FCC unit cell are ?Space for Rough Work


E-21/281001CJA545120002

PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 18)




1. If ( )( )+= -

+sin9A sin A tan k 1 Acos9A cos A (where defined), then k is

(A) 1 (B) 5 (C) 6 (D) 92. The expression 2x2 + 4xy – ky2 + 4x + 2y – 1 can be resolved into two linear factor, then the

value of k is

(A) -53 (B)

53 (C)

13 (D) -

13

3. If cosx + cosy =32 and

5cos 2x cos 2y4

+ = , then cosx cosy is -

(A) 58

- (B) 5

16- (C)

58 (D)

516



E-22/28 1001CJA545120002


4. The minimum value of + +12 18 xyx y , where x,y > 0 is

(A) 12 3 (B) 24 3 (C) 18 (D) 55. If the sum of the roots of quadratic equation ax2 + bx + c = 0 is equal to sum of the squares of

their reciprocals, then +2

2b bcac a

is

(A) 2 (B) –2 (C) 1 (D) –16. Sum of solutions of equation sinx + sin2x + sin3x + sin4x = 0 in [0,p] is

(A) p39

10 (B) p6

5 (C) p32 (D)

p2710


E-23/281001CJA545120002








7. If =

= å6

3 mm

m 0sin x.sin 3x c cos x , where c0,c1,c2,c3,c4,c5,c6 are constant, then

(A) c0 + c2 + c4 + c5 = 0 (B) c1 + c0 + c2 = 5(C) c0 + c2 + c4 + c5 = 1 (D) c0 + c6 = 10

8. Let l1,l2 are the values of x satisfying the equation sinx = cos2x where 0 < l1,l2 < p then

(A) p

l - l =2 123 (B) p

l l =2

1 2536

(C) l + l = p1 2 (D) p

l + l =1 243


E-24/28 1001CJA545120002

9. If equations ax2 – (a + b)x + b = 0 & bx2 + (b – c)x – c = 0 has exactly one root in common{a, b, c ¹ 0}, then which of the following can be correct -(A) b2 = ac (B) –a = b ¹ c (C) b = a ¹ c (D) –a ¹ b = c

10. For the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x

(A) Number of principal solutions are 4 (B) General solution is n , n I2 8p p

+ Î

(C) General solution is n , n I8p

p + Î (D) Number of principal solutions are exactly 2.

11. If A + B + C = p then which of the following is/are true(A) tanA + tanB + tanC = tanA tanB tanC

(B) + + =A B B C C Atan tan tan tan tan tan 12 2 2 2 2 2

(C) sin2A + sin2B + sin2C = 4sinA sinB sinC(D) cotA cotB + cotB cotC + cotC cotA = 1

12. If p

q =8 , then ( ) ( )q - q -2 24 cos 3 4 cos 3 3 is greater than or equals to

(A) 1 (B) 0.5 (C) 2 (D) 3Space for Rough Work


E-25/281001CJA545120002



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 A, G, H are the arithmetic, geometric a.nd harmonic means of two positive real numbers

& if A = 3H, then 2

2

AG

is equal to

2. If sin4x + cos4x = sinx cosx, then the values of x are of the form ( ) p+ Îkn 1 ,n I

k , where k5 is

3. Given the system of equation 2x4 = y4 + z4, xyz = 64 where logyx, logzy, logxz are in G.P(where x,y,z are positive real numbers other than 1), then the value of (x + y – z) is



E-26/28 1001CJA545120002

4. Let a cuboid is having length, width and height as log23, log35, log57. If the volume of cuboid

is l, then 228

l

is

5. For a series, let Tr = r2 + 3r then 10S1100 (Where Sn represent sum of n terms)

6. If k denotes the number of integral value(s) of 'a' for which roots of x2 – (2a + 5)x + a2 + 5a = 0

lies in [–1, 7] then k

10 is ______


E-27/28

ALLENSpace for Rough Work

1001CJA545120002

Leader Live Phase-(I + II) & Leader (On Line + Off Line) Phase-(I + II)/08-11-2020/Paper-1

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. .............................................


Your Target is to secure Good Rank in JEE 2021E-28/28

ALLEN

1001CJA545120002

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 6 corresponding to — of the other 18

the correct option bubbles is casesis darkened darkened

+4 +1 0 –2One or more If only the bubble(s) For darkening a bubble If none In all

I(ii) correct 6 corresponding corresponding to each of the other 24option(s) to all the correct correct option, provided bubbles is casesoption(s) is(are) NO incorrect option darkened

darkened darkened+4 0

Numerical If only the bubble In all II Value Type 6 corresponding — other — 24

(Up to second to correct answer casesdecimal place) is darkened

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lkekU; %

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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 djsaA

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- esa dgha vkSj u fy[ks aA 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

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

fujh

{kd

ds v

uqns'k

ksa d

s fcu

k eqgj

sa u r

ksM+s

Time : 3 Hours Maximum Marks : 198

Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

JEE(Advanced)

08-11-2020

PAPER-1

Hin

di )1001CJA545120002)(1001CJA545120002)

REVIEW TEST

JEE(Main+Advanced) : LEADER LIVE PHASE-(I + II) & LEADER (ON LINE + OFF LINE) PHASE-(I + II)

H-2/28

ALLEN

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

1cm

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

1001CJA545120002



H-3/281001CJA545120002

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

Hkkx-1 : HkkSfrd foKku[k.M–I(i) : (vf/kdre vad : 18)

� bl [k.M esa N% 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 yM+dk ,d dqlhZ ij cSBk gS tks fdlh ?k"kZ.kjfgr f?kjuh ij ls xqtjus okyh jLlh ls tqM+h gSA yM+dk jLlh ds [kqys fljs

dks 250 N cy ls [khaprk gSA yM+ds dk Hkkj 320 N rFkk dqlhZ dk Hkkj 160 N gSA yM+dk Rofjr gks jgk gSA yM+ds ij

dqlhZ dh lhV }kjk yxk;k x;k yxHkx cy gksxk%&

(A) 83 N (B) 70 N (C) 13 N (D) 57 N



H-4/28 1001CJA545120002


2. fp= esa BAC ,d n`<+ fLFkj [kqjnjk rkj gS rFkk dks.k BAC, 60º gSA æO;eku m okyh nks ,d tSlh oy; P rFkk Q ewy

yEckbZ 2a okyh ,d gYdh izR;kLFk jLlh ls tqM+h gq;h gS ftldk cy fu;rkad mga gSA tc PA = AQ = 3a gks rks P

rFkk Q lkE;koLFkk esa gksrs gSA oy; rFkk rkj ds e/; U;wure ?k"kZ.k xq.kkad m gSA m + 3 dk eku gksxk %&

3a

CB

30°30°3a

A

QP

(A) 2 (B) 3 (C) 4 (D) 73. ,d O;fä rj.krky esa 2 m/s ls rSj ldrk gSA og fdlh 100 m pkSM+h unh esa ,d fdukjs ls rSjuk izkjEHk djrk gS rFkk

ty ds lkis{k unh izokg osx ds yEcor~ rSjrk gSA ;fn nwljs fdukjs ij igq¡pus ij mldk viokg 40 m gks rks unh izokg

osx Kkr dhft, %&(A) 0.4 m/s (B) 0.8 m/s (C) 0.6 m/s (D) 2 m/s


H-5/281001CJA545120002


4. iznf'kZr nks fLizax rqyk S1 o S2 ds ikB~;kad Øe'k% 60 kg o 30 kg gSA fLizax S1 foLrkfjr fLFkfr esa gS rFkk fLizax S2

lEihfM+r fLFkfr esa gSA fy¶V 10 m/s2 Roj.k ls Åij dh vksj Rofjr gks jgh gSA fy¶V ds lkis{k æO;eku fLFkj gSA CykWd

dk æO;eku gksxk%&

m

S1

S2

10 m/s2

(A) 45 kg (B) 30 kg (C) 60 kg (D) 15 kg5. ,d d.k 1m f=T;k ds o`rh; iFk ij fLFkjkoLFkk ls izkjEHk gksdj o`Ùkh; xfr djrk gSA fdlh le; t ij d.k dk

Li'kZjs[kh; Roj.k at = t ms–2 fn;k tkrk gSA t = 2 sec ij d.k dk f=T;h; Roj.k gksxk %&(A) 1 ms–2 (B) 2 ms–2 (C) 0.5 ms–2 (D) 4 ms–2

6. ,d ç{ksI; dks {kSfrt ds lkFk dks.k q ij pky u ls QSadk tkrk gSA tc osx {kSfrt ds lkFk dks.k 2qæ ö

ç ÷è ø

cukrk gS rks oØrk

f=T;k gksxh :

(A)2 2

3u cos

2g cos ( / 2)q

q (B) 2 2

3u sin

g sin ( / 2)q

q (C) 2 2

3u cos

g cos ( / 2)q

q (D) 2 2

32u sin

g sin ( / 2)q

q


H-6/28 1001CJA545120002

[k.M -I(ii) : (vf/kdre vad : 24)� bl [kaM esa N% 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, nksuksa 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 : –2 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 xyrgS_ rks dsoy lHkh rhu lgh fodYiksa dk p;u djus ij gh +4 vad feysaxsaA 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), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

7. fuEu esa ls dkSulh oLrq Rofjr gks jgh gS\

(A) vkxs ihNs >wyrk gqvk yksyd

(B) fdlh o`Ùkkdkj iFk ds lkis{k fu;r pky cuk;s j[kdj nkSM+us okyk /kkod

(C) urry ij uhps yq<+drh gqbZ ,d xsan

(D) ljy js[kk esa vf/kdre pky ls xfr'khy ,d f[kykSuk dkj



H-7/281001CJA545120002

8. fp=kuqlkj ,d iz{ksI; fdlh urry ls yEcor~ :i ls Vdjkrk gSA fuEu esa ls dkSulk fodYi lgh gS\

aO

(A) fcUnq O dks iz{ksI; ls tksM+us okys fLFkfr lfn'k dk ifjek.k dHkh ugh ?kVrk gSA

(B) {kSfrt ds Åij mPpre fcUnq rd igq¡pus esa yxk le; mMM~;u dky ds vk/ks ls vf/kd gksrk gSA

(C) iz{ksI; xfr ds nkSjku rkR{kf.kd osx dHkh Hkh izkjfEHkd osx lfn'k ds yEcor~ ugha gks ldrk gSA

(D) urry lss VDdj ds i'pkr~ iz{ksI; dHkh Hkh Å/okZ/kj fn'kk esa xfr ugh dj ldrk gSA

9. {kSfrt /kjkry ij ,d&nwljs ls 70 m dh nwjh ij fLFkr nks d.kksa dks fLFkfr 'A' rFkk 'B' ls ,dlkFk iz{ksfir fd;k tkrk gSA

;fn nksuksa d.k e/; ok;q esa tkdj Vdjkrs gks rks lgh dFku@dFkuksa dks pqfu;s%&

A B45°

50m/s 40 2 m/sÖ

q

70m

(A) ;s vkil esa Vdjk;sxsa ;fn ‘A’ ds fy;s iz{ksi.k dks.k 37° gksA

(B) d.k q ds fdlh Hkh eku ds fy, ugha Vdjk;sxsaA

(C) d.k iz{ksi.k ds i'pkr~ t = 1 sec ij Vdjk;sxsaA

(D) ;s vkil esa Vdjk;sxsa ;fn ‘A’ ds fy;s iz{ksi.k dks.k 53° gksA



H-8/28 1001CJA545120002

10. iznf'kZr O;oLFkk esa lHkh lrgs ?k"kZ.kjfgr gSaA jLlh dk AB Hkkx {kSfrt rFkk BC Hkkx urry ds lekUrj gSA fudk; dks

fojkekoLFkk ls NksM+k tkrk gSA urry ds vuqfn'k CykWd dh xfr ds nkSjku m o M ds e/; vfHkyEc laidZ cy dk eku

'kwU; gS] ost dk Roj.k a rFkk jLlh esa ruko T gS rc %&

mC

BA

Mq

(A) T = (m + M) a (B) ( )MaT

1 cos=

- q(C) a = g cot q (D) T = ma + mg sin q

11. ,d dkj [kqjnjh {kSfrt lrg ij fcuk fQlys o`Ùkkdkj iFk esa okekorZ fn'kk esa xfr dj jgh gSA iFk dk Åij ls ns[kk x;k

n`'; fp= esa n'kkZ;k x;k gSA rv rFkk ra le; 't' ij osx rFkk Roj.k lfn'k dks n'kkZrs gSA tc dkj fcUnq P ij gksrh gS rks uhps

fn;s x;s fodYiksa esa of.kZr fofHkUu fLFkfr;ksa ds fy;s ?k"kZ.k cy dh fn'kk rhjksa }kjk fp= esa n'kkZ;h x;h gSA lgh

fodYi@fodYiksa dks pqfu;s%&

r

5

12

4

3

v®

P

Top view of the circular track

(A) ;fn × =r ra v 0 gks rks ?k"kZ.k cy rhj 3 ds vuqfn'k yxsxkA

(B) ;fn × >r ra v 0 gks rks ?k"kZ.k cy rhj 2 ds vuqfn'k yxsxkA

(C) ;fn × <r ra v 0 gks rks ?k"kZ.k cy rhj 5 ds vuqfn'k yxsxkA

(D) ;fn × <r ra v 0 gks rks ?k"kZ.k cy rhj 4 ds vuqfn'k yxsxkA



H-9/281001CJA545120002

12. ,d CykWd A, osx v ls r[rs B ij xfr djrk gS rFkk r[rk 2v osx ls xfr'khy gSA ;fn /kjkry fpduk gks rFkk A o Bds e/; ?k"kZ.k fo|eku gks rks (Wf = ?k"kZ.k }kjk CykWd A rFkk CykWd B ij fd;k x;k dqy dk;Z gS ; (Wf)A rFkk (Wf)B

?k"kZ.k }kjk Øe'k% CykWd A rFkk B ij fd;s x;s dk;Z gSa)

AB

v

2v

(A) (Wf)A dk eku /kukRed gSA

(B) (Wf)B dk eku ½.kkRed gSA

(C) Wf = WfA + WfB ½.kkRed gSA

(D) Wf funsZ'k ra= ij fuHkZj ugha djrk gSA



H-10/28 1001CJA545120002

[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

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'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d d.k js[kh; :i ls xfr djrk gSA fp= esa a-s oØ n'kkZ;k x;k gSA d.k dk izkjfEHkd osx 10m/s gSA d.k ds 30m nwjh

r; djus ds i'pkr~ d.k dk osx X m/s gS rks X/10 dk eku Kkr dhft;sA

15 30

10

a(m/s )2

s(m)



H-11/281001CJA545120002

2. ,d uko fcUnq A ls js[kk AB ds lkFk dks.k a = 30º cukrs gq;s osx brVr

ls pyuk izkjEHk djds unh dks ikj djrh gSA

fp=kuqlkj unh izokg dk osx rVr

gSA js[kk AC js[kk AB ds lkFk dks.k b = 60º cukrh gSA ;fn uko lnSo js[kk AC ij

jgrh gS rks br

r

VV

dk eku Kkr dhft;sA

Vr

CB

A

Vbr

3. çnf'kZr fp= esa tc fLizax bldh ewy yEckbZ ij gksrh gS rks fudk; dks fojkekoLFkk ls NksM+k tkrk gSA fLizax fu;rkad

100N/m gS rFkk izR;sd CykWd dk æO;eku 10kg gSA CykWd A dk osx 'v' fLFkfr x ds Qyu ds :i esa v2 = ax – bx2

ds vuqlkj fn;k tkrk gSA a/b dk eku Kkr dhft;sA (g =10 m/s2)



H-12/28 1001CJA545120002

4. fdlh ry esa xfr'khy ,d d.k ij F = 10 N fu;r ifjek.k dk ,d cy bl izdkj yxk;k tkrk gS fd ;g fi.M ds osx

( )v v v 5m / s= =r r

ds yEcor~ jgrk gS rFkk cy dh fn'kk lnSo ,d fLFkj fcUnq dh vksj gksrh gSA d.k }kjk S = 10m

nwjh r; djus ij d.k ds osx lfn'k }kjk cuk;k x;k dks.k (jsfM;u esa) gksxk : (d.k dk æO;eku m = 2kg gSA)

5. ,d d.k lehdj.k x = 10 sin 2pt, y = 10 cos 2pt ds vuqlkj xfr djrk gSA t = 0 ls t = 5 sec esa dqy cy }kjk

fd;k x;k dk;Z Kkr dhft;sA



H-13/281001CJA545120002

6. ,d CykWd dk æO;eku M gS rFkk ,d lrg ij fojkekoLFkk eas gS ftlds fy, LFkSfrd rFkk xfrt ?k"kZ.k xq.kkad Øe'k%

ms rFkk mk gSA dscy ij ,d cy F = kt2 yxk;k tkrk gSA t = 2 sec ij CykWd dk osx v m/s gSA 23/v dk eku Kkr

dhft;sA fn;k gS % M = 24 kg, k = 60 N/s2, ms = 0.5, mk = 0.4

FM



H-14/28 1001CJA545120002

Hkkx-2 : jlk;u foKku[k.M–I(i) : (vf/kdre vad : 18)






1. vEyh; lkeF;Z ds fy, lgh Øe gS :

(A)NO2

COOH>

Cl

COOH>

CN

COOH

(B) Cl3C—COOH > CH3COOH > Br3C—COOH(C) CH3—COOH < CH2 = CH—COOH < HCºC—COOH(D) mijksDr lHkh

2. dkcZ½.kk;u LFkkf;Ro ds fy, lgh Øe gS :

(A) > >

1 1 1

(B) H CCH3

CH3

< CH3 CCH3

CH3

1 1

(C) CH2 < CH2

1 1(D) >

1 1

3. fuEu v.kqvksa ds fy,] ca/k dks.k dk xyr Øe dkSu lk gS ?(A) PF3 > PH3 (B) NH3 > NF3

(C) CH3 CH3

O OCH3 Ha b> (D) 2

3CO - > CO2 > 4NH+



H-15/281001CJA545120002


4. fuEu esa ls dkSu lh Lih'kht] T;kfefr rFkk vkÏfr ds vk/kkj ij vU; ls vyx gS \(A) XeO3F2 (B) IF7 (C) 2

6SiF - (D) 5XeF1

5. ,d nzo foy;u esa nks nzo A rFkk B mifLFkr gS];fn foy;u dk ok"i nkc 400 mm Hg gSA fn;k gS

PA0 = 600 mmHg ; PB

0 = 300 mmHg rks vklfor esa A dk eksy izHkkt gS

(A) 1/2 (B) 1/3 (C) 2/3 (D) 1/4

6. ,d rRo tks fcc izdkj dh lajpuk j[krk gS] dh lSy dh fdukjs dh yECkkbZ 'a' pm gSA rRo dk ?kuRo

r gm/cc gS] rks rRo ds W gm esa fdruh bdkbZ lSYk mifLFkr gS \ [NA = vkoksxknzks la[;k]

(A) 30

3W 10a 4

´´ r ´ (B) 30

3W 10

a´

´ r (C) 30

3A

W 10N( 2 a)

´r (D)

36

3A

W 10a N

+´r


H-16/28 1001CJA545120002




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



� 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 fodYiksa dk p;u djus ij gh +4 vad feysaxsaA 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), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

7. xyr ; qXe dk p;u dhft;sA

(A) ,durk{k | a ¹ b ¹ c | Qyd dsfUnzr

(B) f=lerk{k (Rhombohedral) | a = b = c | vk|

(C) "kVdks.kh; | a = b = 90, g = 120º | dk; dsfUnzr

(D) fo"keyack{k (Orthorhombic) | a = b ¹ c | vk|

8. vfHkfØ;k A + 3B ® C ds fy,] lgh dFku dk p;u dhft;s

(A) d[C] d[A]dt dt

-=

(B) 3d[C] d[B]

dt dt= -

(C) nj fu;e r = k [A][B]3 gksuk gh pkfg,(D) vfHkfØ;k dh nj dh bdkbZ;k¡ vfHkfØ;k dh dksfV ls Lora= gksrh gS


H-17/281001CJA545120002

9. fuEu esa ls dkSu ls dFku xyr gSa&(A) HCN esa s rFkk p ca/k vuqikr 1 : 1 gS(B) ca/k ÅtkZ dk Øe : Cl2 > F2 > Br2 > I2

(C) PbF4 rFkk BiF5 vfLrRo ugha j[krs gSa(D) PH3 dh rqyuk esa NH3 vPNk yqbZl {kkj gS

10. fuEu esa ls dkSuls dFku lgh gSa&(A) PCl5(s) ® /kuk;fud Hkkx (sp3 ladfjr) + ½.kk;fud Hkkx (sp3d2 ladfjr)(B) PBr5(s) ® /kuk;fud Hkkx (sp3 ladfjr) + ½.kk;fud Hkkx (mRÏ"V xSl foU;kl)(C) I2Cl6(l) ® /kuk;fud Hkkx (sp3 ladfjr)+ ½.kk;fud Hkkx (sp3d2 ladfjr)

(D) N2O5(s) ® /kuk;fud Hkkx (js[kh;) + ½.kk;fud Hkkx (f=Hkqth; leryh;)

11. LFkkf;Ro dk lgh Øe igpkfu;s&

(A) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

(B) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

– – – –

(C) CH < CH CH < CH CH CH < CH C CH3 3 2 3 3 3 3— — — — —

CH3

+ + + +

(D) bueas ls dksbZ ughadPps dk;Z ds fy, LFkku


H-18/28 1001CJA545120002

12. lgh Øe igpkfu;s&

(A) H—F < H—Cl < H—Br < H—I vEyh; lkeF;Z

(B) CH4 < NH3 < H2O < H—F vEyh; lkeF;Z

(C)

NH2

<

NH2

{kkjh; lkeF;Z

(D)

Cl

F

OBr–

>F

Cl

O–

Br

>Cl

Br

O–

FLFkkf;Ro Øe



H-19/281001CJA545120002




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. la;qXeu esa fdrus p-bysDVªkWu lfEefyr gSA

2.OH

ds fy, fdruh vuquknh lajpuk,sas lEHkkfor gS



H-20/28 1001CJA545120002

3. fuEu esa ls , sls v.kqvksa dh la[;k crkbZ;s ftuesa pp-pp rFkk pp-dp, nksuksa izdkj ds ca/k mifLFkr gSaCOCl2, SO2, SO3, B3N3H6, XeO4, POCl3

4. fuEu eas ls , slh jklk;fud Lih'kht dh la[;k crkbZ;s ftudk vfLrRo ugha gSNCl5, SI6, TlI3, FeI3, CuI2, PbI4, BiI5, NOF3

5. nzo foy;u esa leku nzO;eku ds A rFkk B mifLFkr gS rks ok"i }kjk mRiUu dqy nkc X torr gS] X/6 dk eku

crkb;sA [A = 80 gm/mol, B = 120 gm/mol]

Pr(torr)

30

0100

1000

PT

PB PA

A dk nzO;eku %

40

Pr(torr)

B dk nzO;eku %

6. FCC bdkbZ lsy esa 3rd fudVre iM+kSlh ijek.kq dh la[;k fdruh gS?dPps dk;Z ds fy, LFkku


H-21/281001CJA545120002

Hkkx-3 : xf.kr[k.M–I(i) : (vf/kdre vad : 18)






1. ;fn ( )( )+= -

+sin9A sin A tan k 1 Acos9A cos A (tgk¡ ifjHkkf"kr) gks] rks k dk eku gksxk

(A) 1 (B) 5 (C) 6 (D) 92. O;atd 2x2 + 4xy – ky2 + 4x + 2y – 1 dks nks js[kh; xq.ku[k.Mksa esa foHkDr fd;k tk ldrk gS] rks k dk eku gksxk

(A) -53 (B)

53 (C)

13 (D) -

13

3. ;fn cosx + cosy =32 rFkk

5cos 2x cos 2y4

+ = , rks cosx cosy dk eku gksxk -

(A) 58

- (B) 5

16- (C)

58 (D)

516



H-22/28 1001CJA545120002


4. + +12 18 xyx y , (tgk¡ x,y > 0) dk U;wure eku gksxk

(A) 12 3 (B) 24 3 (C) 18 (D) 55. ;fn f}?kkr lehdj.k ax2 + bx + c = 0 ds ewyksa dk ;ksxQy] buds O;qRØeksa ds oxk±s ds ;ksxQy ds cjkcj gks] rks

+2

2b bcac a

dk eku gksxk

(A) 2 (B) –2 (C) 1 (D) –16. vUrjky [0,p] esa lehdj.k sinx + sin2x + sin3x + sin4x = 0 ds gyksa dk ;ksxQy gksxk

(A) p39

10 (B) p6

5 (C) p32 (D)

p2710


H-23/281001CJA545120002




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



� 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 fodYiksa dk p;u djus ij gh +4 vad feysaxsaA 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 ), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa ) dks pquk x;k gks ;k u pquk x;k gksA

7. ;fn =

= å6

3 mm

m 0sin x.sin 3x c cos x , tgk¡ c0,c1,c2,c3,c4,c5,c6 vpj gks] rks

(A) c0 + c2 + c4 + c5 = 0 (B) c1 + c0 + c2 = 5(C) c0 + c2 + c4 + c5 = 1 (D) c0 + c6 = 10

8. ekuk x ds eku l1,l2 gS tks lehdj.k sinx = cos2x, (tgk¡ 0 < l1,l2 < p) dks larq"V djrs gS] rks

(A) p

l - l =2 123 (B) p

l l =2

1 2536

(C) l + l = p1 2 (D) p

l + l =1 243


H-24/28 1001CJA545120002

9. ;fn lehdj.k ax2 – (a + b)x + b = 0 rFkk bx2 + (b – c)x – c = 0, {a, b, c ¹ 0} dk Bhd ,d ewy mHk;fu"B gks] rks

fuEu esa ls dkSulk fodYi lgh gks ldrk gS \(A) b2 = ac (B) –a = b ¹ c (C) b = a ¹ c (D) –a ¹ b = c

10. lehdj.k sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x ds fy;s&

(A) eq[; gyksa dh la[;k 4 gksxhA (B) O;kid gy n , n I2 8p p

+ Î gksxkA

(C) O;kid gy n , n I8p

p + Î gksxkA (D) eq[; gyksa dh la[;k Bhd 2 gksxhA

11. ;fn A + B + C = p gks] rks fuEu esa ls dkSulk@dkSuls lR; gksxk@gksaxs\(A) tanA + tanB + tanC = tanA tanB tanC

(B) + + =A B B C C Atan tan tan tan tan tan 12 2 2 2 2 2

(C) sin2A + sin2B + sin2C = 4sinA sinB sinC(D) cotA cotB + cotB cotC + cotC cotA = 1

12. ;fn p

q =8 gks] rks ( ) ( )q - q -2 24 cos 3 4 cos 3 3 dk eku fuEu ls vf/kd ;k cjkcj gksxk

(A) 1 (B) 0.5 (C) 2 (D) 3dPps dk;Z ds fy, LFkku


H-25/281001CJA545120002




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 A, G, H nks /kukRed okLrfod la[;kvksa ds lekUrj] xq.kksÙkj rFkk gjkRed ek/; gks rFkk A = 3H gks] rks 2

2

AG

dk eku gksxk

2. ;fn sin4x + cos4x = sinx cosx gks] rks x dk eku tks ( ) p+ Îkn 1 ,n I

k ds :i esa gS] tgk¡ k5 dk eku gksxk

3. fn;k x;k lehdj.k fudk; 2x4 = y4 + z4, xyz = 64 tgk¡ logyx, logzy, logxz xq.kksÙkj Js.kh esa gS

(tgk¡ x,y ,oa z, 1 ds vfrfjDr /kukRed okLrfod la[;k;sa gSa), rks (x + y – z) dk eku gksxk



H-26/28 1001CJA545120002

4. ekuk ,d ?kukHk ftldh yEckbZ] pkSM+kbZ rFkk Å¡pkbZ Øe'k% log23, log35, log57 ds :i eas gSA;fn ?kukHk dk vk;ru

l gks] rks 228

l

dk eku gksxk

5. Js<+h ds fy;s ekuk Tr = r2 + 3r gks] rks 10S1100 (tgk¡ Sn, n inksa ds ;ksxQy dks n'kkZrk gS) gksxk

6. ;fn 'a' ds iw.kk±d ekuksa dh la[;k dks k ls n'kkZrs gS ftlds fy;s lehdj.k x2 – (2a + 5)x + a2 + 5a = 0 ds ewy

vUrjky [–1, 7] esa fLFkr gks] rks k

10 dk eku gksxk


H-27/28

ALLENdPps dk;Z ds fy, LFkku

1001CJA545120002

Leader Live Phase-(I + II) & Leader (On Line + Off Line) Phase-(I + II)/08-11-2020/Paper-1

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 ;k stuk %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


Your Target is to secure Good Rank in JEE 2021H-28/28

ALLEN

1001CJA545120002

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 iwjhrjg tk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksMnksuks a leku gSaA

____________________________

fujh{kd ds gLrk{kj

ijh{kkFkhZ dk uke ................................................................................................

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

[k.M iz'u dk iz'uks a oxkZuqlkj izR;sd iz'u ds vad [k.M esaizdkj 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 6 ;fn flQZ lgh fodYi ds — ;fn fdlh Hkh vU; lHkh 18vuq:Ik cqycqys dks cqycqys dks dkyk ifjfLFkfr;ksa

dkyk fd;k gS ugha fd;k gS esa,d ;k ,d ls +4 +1 0 –2

I(ii) vf/kd lgh ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i ;fn fdlh Hkh vU; lHkhfodYi 6 (fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] cqycqys dks ifjfLFkfr;ksa 24

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 II izdkj ;fn flQZ lgh mÙkj ds vU; lHkh

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

LFkku rd) dkyk fd;k gS esa

PAPER-1 - Doubtion

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