(PANINI-eSANKALP022)

IIT – JEE, 2021

PAPER 1 (PHASE –IV)
Time: 3 Hours Maximum Marks: 177

INSTRUCTIONS

A. Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections.
1. Section 1 contains 09 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which only
ONE option is correct.
2. Section 2 contains 2 paragraphs each describing theory, experiment and data etc. Four questions relate to two
paragraphs with two questions on each paragraph. Each question pertaining to a particular passage should have only one
correct answers among the four given choices (A), (B), (C) and (D).
3. Section 3: contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which One
Or More Than One are correct.
B. Marking Scheme
7. For each questions in Section 1 and 2: you will be awarded 3 marks if you darken the bubble corresponding to the
correct answer only and (–1) mark for wrong answer and 0 marks if no bubbles are darkened.
8. For each question Section 3: marks will be awarded in one of the following categories:
Full Marks : + 4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.
Partial Marks : + 1 For darkening a bubble corresponding to each correct option, provided NO
incorrect option is darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : – 2 In all other cases.
For example, if [A], [C] and [D] are all the correct options for a question, darkening all these three will get +4 marks;
darkening only [A] and [D] will get +2 marks; and darkening [A] and [B] will get -2 marks, as a wrong option is also
darkened.

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-2

PART I: PHYSICS
SECTION – 1 : (Only One Option Correct Type)
This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. The total energy of a particle executing S.H.M. is proportional to

(A) Displacement from equilibrium position (B) Frequency of oscillation
(C) Velocity in equilibrium position (D) Square of amplitude of motion
2. A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.)
and total energy (T.E.) are measured as a function of displacement x. Which of the following
statements is true
(A) P.E. is maximum when x = 0 (B) K.E. is maximum when x = 0
(C) T.E. is zero when x = 0 (D) K.E. is maximum when x is maximum
3. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the
displacement of the wire is y1  A sin( x / L)sin t and energy is E1, and in another experiment
its displacement is y 2  A sin(2x / L)sin 2t and energy is E2. Then
(A) E2  E1 (B) E2  2E1 (C) E2  4E1 (D) E2  16E1
4. One mole of hydrogen and one mole of oxygen are at the same temperature. Therefore the r.m.s.
velocities of hydrogen and oxygen molecules are in the ratio.
(A) 1 : 1 (B) 1 : 4 (C) 4 : 1 (D) 16 : 1
5. Graph between velocity and displacement of a particle, executing SHM is
(A) A straight line (B) A parabola (C) A hyperbola (D) An ellipse
6. The displacement of a particle varies according to the relation x  4(cos t  sin t) . The amplitude
of the particle is
(A) 8 (B) – 4 (C) 4 (D) 4 2

7. An ideal monatomic gas is taken round the cycle ABCDA as shown in 2P, V 2P, 2V
the figure. The work done during the cycle is B C
(A) PV/2 (B) PV P
A D
(C) 2PV (D) zero
P, V P, 2V
V

8. A particle is executing SHM. Then the graph of acceleration as a function of displacement is

(A) A straight line (B) A circle (C) An ellipse (D) A hyperbola

9. If a spring has time period T, and is cut into n equal parts, then the time period of each part will
be
(A) T n (B) T / n (C) nT (D) T

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SECTION – 2: Comprehension Type (Only One Option Correct)

This section contains 2 Paragraphs, each describing theory, experiments, data etc. 4 Questions relate
to the 2 Paragraphs with 2 Questions on Each Paragraph. Each question has Only One Correct
Answer among the four given options (A), (B), (C) and (D).
Paragraph for Question Nos. 10 to 11
When more than one force (say two forces) acts on a system it might produce more than one
S.H.M’s. The combination may form another SHM depending on the direction of such S.H.M, the
amplitude will vary. The phase difference between the two also has a role to play in deciding the
resultant amplitude. However when the supper imposed S.H.M’s are in perpendicular direction,
the pattern may change not only with phase but also with frequencies. When different oscillating
systems are connected, there can be an influence of one on another.

10. When two S.H.M’s in the same direction with amplitude A1 and A2 are super-imposed, the
resultant amplitude will be
(A) | A1  A2 | always (B) | A1  A2 | for  = 

(C) | A1  A2 | for  = 0 (D) between | A1  A2 | and A1  A2  if 0    

11.  
If Y1  5 sint  and Y 2  5 3 sin t  cos t are two S.H.M’s the ratio of their amplitude is


(A) 1 : 3 (B) 1 : 3 (C) 1 : 2 (D) 1 : cos 
6

Paragraph for Question Nos. 12 and 13

When two pulses are allowed to pass through a string in opposite directions with the same speed the
shape of the string varies. Each pulse pass the overlap region as if the other pulse was not at all there.
Super position principle can be applied to deal with two waves in opposite direction. If two wave pulses
are identical in shape except that one is inverted with respect to another, at some extent displacement will
be zero, but not their velocities. When reflection takes place from a denser medium’s surface or a rigid
body, there will be phase difference of  radians. The corresponding path and time difference can be
analysed.
12. If the wave travelling in a string, given by Y  A sint  kx  is reflected at x = 0 from a fixed
end, the equation of the reflected wave is
(A) Y  A sint  kx  (B) Y   A sint  kx 

(C) Y   A sint  kx  (D) Y   A sin t  kx  

13. In reflected wave, which quantities will have same magnitude

(A) Velocity (B) amplitude (C) wavelength (D) all of above

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-4

SECTION – 3: (One or More Than One Options Correct Type)

This section contains 5 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.
14. A wave disturbance in a medium is described by
 
y(x, t)  0.02cos  50  t    cos(10 x) , where x and y are in metres and t in seconds
 2

(A) A displacement node occurs at x = 0.15 m (B) An antinode occurs at x = 0.3 m

(C) The wavelength of the wave is 0.2 m (D) The speed of the wave is 5.0 m/s

15. The (x, y) coordinates of the corners of a square plate are (0, 0), (L, 0) (L, L) and (0, L). The
edges of the plate are clamped and transverse standing waves are set up in it. If u(x, y) denotes
the displacement of the plate at the point (x, y) at some instant of time, the possible expression(s)
for u is(are) (a = positive constant)

x y x y
(A) a cos cos (B) a sin sin
2L 2L L L

x 2 y 2 x y
(C) a sin sin (D) a cos cos
L L L L

16. A black body at temperature T is lying inside a closed chamber which is

at temperature T0 initially. Now a small hole is made at the top of the T0
T
chamber and sun rays are allowed to fall from it. Then
(A) Black body will absorb more radiation
(B) Black body will absorb less amount of radiation
(C) Black body will emit more energy.
(D) Black body will emit energy equal to energy absorbed by it after steady state is reached
For temperature to remain constant, energy absorbed should be equal to energy emitted.

17. At ordinary temperatures, the molecules of an ideal gas have only translational and rotational
kinetic energies. At higher temperatures, they may also have vibrational energy. As a result, at
higher temperatures
(A) Cv = 3R/2 for monatomic gas (B) Cv > 3R/2 for monoatomic gas
(C) Cv < 5R/2 for diatomic gas (D) Cv > 5R/2 for diatomic gas

18. When an ideal monoatomic gas is heated at constant pressure, which of the following may be
true
dU 3 dW 2 dU 4
(A)  (B)  (C)  (D) dW + dU = dQ
dQ 5 dQ 5 dQ 5

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 IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-5

PART II: CHEMISTRY

SECTION – 1 : (Only One Option Correct Type)
This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. H2 O,H 
 X
i Hg OAc 
 ii NaBH4  Y
 2

 CH3 3 C  CH  CH2
 3i BH  THF

 ii H O /OH
Z
2 2

The correct statement from the above sequence is:

(A) X and Y are optically active where as Z is inactive
(B) Y and Z are optically active where as X is inactive
(C) X and Z are optically inactive where as Y is active
(D) All X, Y and Z are optically inactive

2. Consider the following organic compound:

H

HO CH3
The correct IUPAC name of the above compound is:
(A) 4-(3-methyl-5-hydroxyphenyl)-3-methylpent-2-enal
(B) 4-(3-hydroxy-5-methylphenyl)-3-methylpent-2-enal
(C) 4-(3-methylphenol)-pent-2-enal
(D) 4-(3-hydoxy-5-methylphenyl)-3,4-dimethylbut-2-enal

3. Which of the following is the most stable carbanion?

CH 2 CH 2 CH 2
CH 2
(B) (C) (D)
(A)

CH3 NO2 Cl
4.
Br2, H2O
Product(s)
NaCl
The product(s) in the above reaction that cannot be formed is:

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-6

Br Br
(A) (B)
Cl Br
Cl Br
(C) (D)
OH OH

5. O
H2C C NH2 (IV)

O
HC C 6H4 NH C CH3
(III)

H2C NH CH CH2 NH2 (I)

(II)
CH3
Arrange the above mentioned nitrogen atoms (I, II, III, IV) of the compound in decreasing order of
basic nature?
(A) I > II > IV > III (B) II > I > IV > III (C) II > I > III > IV (D) I > IV > III > II

6. Major product of the following reaction, is:

CH3

Br2 / FeCl 3
(major product)
(dark)
CH3
CH3 CH3 H2C Br
Br
(B) (C) (D)
(A)

Br
Br

7. Indicate the correct relationship in the following three pairs:

F

Br H H
CH2Br
and
(I) H3C
F
H CH3

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 IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-7

H C 2H 5 Cl H
Cl and H
H
(II) H3C Cl
Cl
CH3
C 2H 5

CH3 CH3

H Br Br H
and
(III) H3C H H CH3

Br Br

I II II

(A) Enantiomeric pair Diasteromeric pair Enantiomeric pair

(B) Identical Enantiomeric pair Enantiomeric pair

(C) Enantiomeric pair Diastereomeric pair Identical

(D) Enantiomeric pair Identical Identical

8. Which of the following pair does not show enantiomerism

Cl Cl Cl Cl
(A) (B)

Cl Cl
Cl Cl
H Br H Cl
(C) H Br (D)
H Cl Cl
Cl H
H Et
Cl Et
H Cl H

+
9. Which of the following compounds is formed in a major quantity when reacted with H ion?
H 2C CH2 CH CH CH2
+
CH3 H
major product

H2 C
CH2

(A) CH CH3 (B)

Me Et
H2 C CH3

(C) Both (A) and (B) (D) CH2CH3

CH3

SECTION – 2: Comprehension Type (Only One Option Correct)

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-8

This section contains 2 Paragraphs, each describing theory, experiments, data etc. 4 Questions relate
to the 2 Paragraphs with 2 Questions on Each Paragraph. Each question has Only One Correct
Answer among the four given options (A), (B), (C) and (D).
Paragraph for Question Nos. 10 to 11

During the practice of isomerism in a class a student Paul was given an exercise by his teacher E.Smith
as following:
“Prepare the structure of organic compounds strictly following given conditions with using only one or
more atoms among carbon (C), hydrogen (H), deuterium (D) and oxygen (O). [MW of C(12), H(1), D(2),
O(16)]
Condition 1: You cannot use more than three similar atoms in a molecule.
Condition 2: The molecule must be saturated.

Now, answer the following questions:

10. If Paul has to draw the structure of an optically active molecule with lowest possible molecular
weight by following above conditions then what would be the molecular weight of that structure?

(A) 46 (B) 47 (C) 48 (D) 49

11. What will be the lowest possible molecular weight for the compound which is optically inactive
prepared with using above conditions by Paul.

(A) 15 (B) 16 (C) 17 (D) 18

Paragraph for Question Nos. 12 to 13

The transformation shown below can be accomplished in three steps.

CH3 O 2N COOH

Br
The steps are as follows;
+3
1. Bromine may be introduced by heating with Br2 and a Fe catalyst.
2. Nitro groups may be introduced by heating with conc. HNO3 and H2SO4.
3. A carboxyl group may be created by KMnO4 oxidation of the methyl group followed by acidification.

There are six possible combinations of these three steps, not all of which will lead to the desired
compound. Consequently, it is important to combine them in an effective manner.

Answer the following questions based on the above information:

12. The sequence in which the steps to be used to accomplish the transformation is
(A) 1, 2, 3 (B) 3, 1, 2 (C) 1, 3, 2 (D) 2, 1, 3

13. If the final product is subjected to further EAS(electrophilic aromatic substitution), the incoming
electrophile will get itself attached to a position
(A) ortho to –NO2 group (B) ortho to –CO2H group and para to –NO2 group
(C) ortho to – Br group (D) ortho to both –CO2H and –NO2 groups

SECTION – 3: (One or More Than One Options Correct Type)

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

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 IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-9

14. Which of the following carbocation can rearrange to get stabilized?


(B) (C) (D)
(A)  


15. Choose the correct statements about following relations.

CCl4
 C  C  + Br 2  C = C  (X)

Br
+
CCl4
C=C + Br2 CC (Y)
Br
+
(A) ‘Y’ is aromatic in character (B) ‘X’ is antiaromatic in character
(C) ‘X’ is much more strained than ‘Y’ (D) ‘Y’ is nonaromatic

16. Which of the following is/are the correct set of compounds in increasing order of pKb value?
(A) NH2 NH (B)
N NH < N NH
<

(C) Cyclohexylamine < aniline < (D) p-chloroaniline < m-chloroaniline <
N-phenyl aniline o-chloroaniline

17. Identify the correct statement/s in the following:

(A) 1,2-dimethyl cyclopentane has four stereo isomers
(B) both cis and trans 1,4-dimethyl cyclohexanes are optically inactive
(C) 1,3-dimethyl cyclohexane has three stereo isomers
(D) 1,2-dimethyl cyclohexane has four resolvable optical isomers at ordinary temperature

18. Which of the following statements is/are correct for cis-1,2-dibromocyclopentane?

(A) It contains two chiral centres, but is optically inactive
(B) It can exist in two enatiomeric forms but cannot be optically active
(C) It is a meso compound
(D) It is with two chiral centres and is optically active

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-10

PART III: MATHEMATICS

SECTION – 1 : (Only One Option Correct Type)
This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.
1. The number of positive integral solution of the equation xyz = 3000 are
(A) 200 (B) 300 (C) 3000 (D) 50

2. If z1  z2 and arg  z1   arg  z2    , then

(A) z1  z2 (B) z1  z2 (C) z1 z2  1 (D) z1   z2

n n
1 r
3. If a n   nC , then  nC equals
r 0 r r 0 r
n n
(A) an (B) an (C) na n (D) n  1a n
2 4
4. The locus of Z which lies in shaded region (excluding the boundaries) is best represented by
 P
(A) z : z  1  2 and arg  z  1   2  1, 2  B
4
A C
 ( 1, 0) (1, 0)
(B) z : z  1  2 and arg  z  1  D
4
 2  1,  2 

(C) z : z  1  2 and arg  z  1 
4

(D) z : z  1  2 and arg  z  1 
4
5. Let d1 , d 2 , ....., d k be all the divisors of a positive integer n including 1 and n. Suppose
1 1 1
d1  d 2  ...  d k  72. Then the value of   .... 
d1 d 2 dk

k2 72 72
(A) is (B) is (C) is
72 k n
(D) cannot be computed from the given information
6. Let n be a natural number and C0, C1,…, Cn be the binomial coefficients. Then
n+1 n n–1
(n+ 2) C0 2 – (n+ 1) C1 2 + n C2 2 + .. is equal to
(A) 4n (B) 4 (C) 2n+4 (D) 4( 1+n)

2 2 1 2a
7. If    , where a, b  N, then the ordered pair (a, b) is
9! 3! 7! 5! 5! b!
(A) (9, 10) (B) (10, 9) (C) (10, 7) (D) (7, 10)

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 IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-11

n  r 1 n r 
 Cr . Cr1 .2r1  is equal to;
8.   
r 1  r1  0


n n n n n n n n
(A) 4 –3 + 1 (B) 4 –3 –1 (C) 4 –3 + 2 (D) 4 –3

2 2 2
9. The ratio in which the sphere x + y + z = 504 divides the line segment AB joining the points
(12, 4, 8) and (27, 9, 18) is given by
(A) 2 : 3 externally (B) 2 : 3 internally (C) 1 : 2 externally (D) None of these

SECTION – 2: Comprehension Type (Only One Option Correct)

This section contains 2 Paragraphs, each describing theory, experiments, data etc. 4 Questions relate
to the 2 Paragraphs with 2 Questions on Each Paragraph. Each question has Only One Correct
Answer among the four given options (A), (B), (C) and (D).
Paragraph for Question Nos. 10 to 11
Let z1 & z 2 be two complex numbers satisfying z  2  z   2  2  0 , where  is non real cube root of
1 3
unity. Also let ‘  ’ be a variable point on the circle z   and 1 ,  2 are values of  such that  is
2 2
maximum and minimum respectively.
On the basis of above information, answer the following questions:
2 2 2 2
10. Value of z1    z 2        2   is

(A) 3 (B) 3 (C) 6 (D) 1

11. Number of complex numbers  such that   1     2  6 is

(A) 4 (B) 3 (C) 2 (D) 0

Paragraph for Question Nos. 12 to 13

100 100  k 1 100  99  100  98  100 100  k  n

Let S k  3 k     3     3k  2     ....   1k    , where  
 0  k   1  k  1  2  k  2   k  0  r
k
1
represents the coefficient of x r in the binomial expansion of 1  x n  n  N and Vk    S k  M 100, k  .
2

100
12. The value of  SkS100 k equals
k 0

 200   200   200   200 

(A)   (B) 2100   (C) 2 200   (D) 2 99  
 100   100   100   99 

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IT-2022-SANKALP022-LOT-A, B&C–PH-IV -PCM-P1-12

13. In an examination, there are 100 true-false type questions. The number of ways in which one
student can make his choice to attempt or not to attempt and then answer the questions is given
by

100 100 100 100

(A)  Sk (B)  Vk (C)  Sk (D)  Vk
k 0 k 0 k 1 k 1

SECTION – 3: (One or More Than One Options Correct Type)

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

14. The number of positive integral solutions of x + y + z + w = 26 s.t

(A) x < y is 1078 (B) x = y is 144 (C) x > y is 2255 (D) x = y is 378

15. If 1,  ,  2 ,...,  n 1 are the nth roots of unity, then  2     2   2  ...  2   n 1  equals

(A) 2n  1 (B)
n
C1  nC2  ...  nCn
1/ 2
(C)  2 n 1C1  2 n 1C2  ...  2 n 1Cn  1 (D) 2
n

16. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C
are separated form one another is :
13 12! 13! 14! 13!
(A) C3 . (B) (C) (D) 11.
5! 3! 2! 5! 3! 3!2! 3! 3!2! 6!

17. A women has 11 close friends. Number of ways in which she can invite 5 of term to dinner, if two
particular of them are not on speaking terms & will not attend together is :
11 24 9 9 9
(A) C5 – C4 (B) C5 + 2 C4 (C) 3 C4 (D) None

18. If z 0 , z1 represent points P, Q on the locus z  1  1 and the line segment PQ subtends an angle


at the point z  1 then z 1 is equal to
2
i
(A) 1  iz 0  1 (B) (C) 1  iz 0  1 (D)  iz 0  1
z 0 1

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