(Batches: e-SANKALP-2325 S1 & T1)

I I T – JEE, 2 3 2 5 Paper Code

100829
(CLASS XI)
Time: 2 Hours Maximum Marks: 126
INSTRUCTIONS
A. General
1. Write your Name, Enrolment number in the space provided on this booklet as soon as you get the paper.
2. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers, and electronic
gadgets of any kind are NOT allowed in the examination hall.
3. Use a ball point pen do darken the bubbles on OMR sheet as your answer besides Name, Enrolment
number, Phase, Paper sequence, Venue, Date along with your signature on OMR sheet.

B. Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of
three sections.
4. Section–1 (01 – 06) contains (06) Multiple Choice Questions which have Only One Correct answer. Each
question will be evaluated according to the following marking scheme.
Full Marks : +3 If only (all) the correct option(s) is (are) chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases
5 Section–2 (07 – 09) contains (03) Multiple Choice Questions which have one or more than one
correct answer. 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 and
both of which are correct;
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.
6. Section–3 (10 – 12) contains (03) Non-Negative Numerical Value Questions, the answer to each
question is a Non-Negative Numerical Value. For each question, enter the correct numerical value
corresponding to the answer and each question carries +4 marks for correct answer. There is no
negative marking.

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-2

PART I : PHYSICS
SECTION – 1 : (Only One Option Correct Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. A car C of mass m is initially at rest on the boat A of mass C

m
M tied to the identical boat B of same, mass M through a M A B M
massless inextensible string as shown in the figure. The car
accelerates from rest to velocity v0 with respect to boat A in
time t0 sec. At time t = t0 the car applies brake and comes to
rest relative to boat in negligible time. Neglect friction
between boat and water find the velocity of boat A just after
applying brake.
Mmv 0 Mmv 0
(A) (B)
(2M  m)(M  m) (M  2m)(M  m)
2Mmv 0
(C) (D) zero
(M  2m)(M  m)

2. Two point particles (2m and m) are connected through 2 m/s


a string of length 4m with separation between them
2m
very slightly less than 4m as shown. They are given m
A B
horizontal velocities as shown and then the system
proceeds under gravity. The string breaks just after the
tension in it reaches maximum value such that particle
B then follows a vertical trajectory. If A and B land
when the separation between them is minimum. Find
the initial height H of the arrangement.
(A) 20 m (B) 39.2 m
(C) 40 m (D) 78.4 m

3. A block of mass m is attached with a spring of force constant k. The k

block is kept on a frictionless plank. Mass of the plank is M and it is P m
also kept on a horizontal frictionless surface. Initially the system is
M
stationary. An impulse P is applied on the block as shown. The
maximum compression in the spring will be Frictionless surface

MP2 mP2
(A) (B)
m(M  m)k M(M  m)k
MP2 mP2
(C) (D)
m2k M2k

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100829)-3

4. The chain has a mass per unit length . Neglect

any friction in the pulley and horizontal surface on y
which chain lies, but coefficient of friction
3 H
between inclined plane and the chain is . A  = 3/4
4
37° F
force F is required to give the open link of the
chain a constant velocity v 0 (  Hg). The =0
magnitude of force when y = H/2 is (assume that
there is no friction between the links of the piled
up portion of chain on the horizontal ground)
 7Hg   3Hg 
(A)   (B)  2 
 2   
 5Hg   Hg 
(C)  . (D)  2 
 2   

5. A ball ‘A’ of mass ‘m’ is placed on a smooth horizontal m

surface. An identical ball ‘B’ of same mass ‘m’ moving in B
the vertical plane with a velocity v0 = 5m/s at an angle 60°
v0
from horizontal collides elastically with the ball ‘A’ as m
shown. Then the velocity of the ball ‘A’ just after collision is 60 A
(A) 1 m/s (B) 2 m/s
(C) 3 m/s (D) 4 m/s
smooth

6. A mass of 3M moving at a speed v collides with a mass of M moving directly towards it, also with
a speed v. If the collision is completely elastic, the total kinetic energy after the collision is K e . If
Ke
the masses stick together, the total kinetic energy after the collision is K s . What is the ratio ?
Ks
1
(A) (B) 4
4
1
(C) (D) 2
2

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-4

SECTION – 2 : (One or More Than One Options Correct Type)

This section contains 3 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

7. In figure, block 1 of mass m1 slides along an x axis

on a frictionless floor at speed 9 m/s. Then it
undergoes a one-dimensional elastic collision with
stationary block 2 of mass m2 = 2m1. Next, block 2
undergoes a one-dimensional elastic collision with
stationary block 3 of mass m3 = 2 m2.
After the collisions are over, which of the following
statement(s) is(are) correct ?
(A) Block 3 has a speed of 4 m/s
(B) The speed of Block 3 is less than the initial speed of block 1
(C) The momentum of block 3 is greater than the initial momentum of block 1
(D) The KE (kinetic energy) of block 3 is less than the initial KE of block 1

8. Ball ‘A’ is in motion with linear momentum ‘P’ while ball ‘B’ is at rest. A B
Masses of the both the balls are same and equal to m. Ball ‘A’
obliquely collide with the ball ‘B’. After the collision, ball ‘A’ moves at P
m m
an angle of 30 with the initial direction while momentum of the ball ‘B’
2
is times of its minimum possible value. Then:
3
(A) after the collision both ball ‘A’ and ‘B’ moves at an angle of 60 with each other.
(B) after the collision both ball ‘A’ and ‘B’ moves at an angle of 90 with each other.
P
(C) after the collision momentum of the ball ‘A’ is is .
3
P2
(D) loss in kinetic energy after the collision is .
6m

9. A ball of mass 2 kg moving horizontally with velocity 10 m/s hits a

wedge of mass 5 kg placed on a horizontal surface as shown in
10 m/s
the figure. Just after collision velocity of wedge is 3.2 m/s. There
is no friction at any contact surface. Then (take sin 37 = 3/5)
(A) Speed of ball just after collision is 210 m/s 53
(B) Impulse applied by ball on the wedge is 20 N-s
(C) Coefficient of restitution for the collision is approximately 0.68
(D) If the time of contact between ball and wedge during collision is 0.05 sec, the average force
exerted by the horizontal surface on the wedge during collision is 260 N.

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100829)-5

SECTION – 3: (Numerical Answer Type)

This section contains THREE 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. xxxxx.xx).

10. Infinite number of bricks are placed one b/2 M/4

over the other as shown in the figure. Each b M a/2 O
succeeding brick having half the length and
a
breadth of its preceding brick and the mass
of each succeeding bricks being (1/4)th of
the preceding one. Taking ‘O’ as the origin,
the x coordinate of centre of mass of the
3a
system of bricks is at . Find the value
k
of k.

M
11. In the shown figure, a particle of mass strikes the block of mass M
10
with velocity v0 and gets attached to it. For what velocity v0 (in ms-1), M
the block B just able to leave the ground ? 10 A
(Given M = 100 gm, K = 880 N/m) K
M

B
2M

12. A spring-block system is resting on a frictionless floor

as shown in the figure. The spring constant is 2.0 N m−1
and the mass of the block is 2.0 kg. Ignore the mass of
the spring. Initially the spring is in an unstretched 2 ms–1
condition. Another block of mass 1.0 kg moving with a 1 kg 2 kg
speed of 2.0 m s−1 collides elastically with the first
block. The collision is such that the 2.0 kg block
does not hit the wall. The distance, in metres, between the two blocks when the spring returns to
its unstretched position for the first time after the collision is _________. ( = 3.14)

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-6

PART II : CHEMISTRY
SECTION – 1 : (Only One Option Correct Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the

following equation: X2  g    2X  g 
The standard reaction Gibbs energy, rG0, of this reaction is positive. At the start of the reaction,
there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is
given by . Thus, equilibrium is the number of moles of X formed at equilibrium. The reaction is
carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R
= 0.083 L bar K1 mol1)
The INCORRECT statement among the following, for this reaction is
(A) Decrease in the total pressure will result in formation of more moles of gaseous X
(B) At the start of the reaction, dissociation of gaseous X2 takes place spontaneously
(C) equilibrium = 0.7
(D) Kc < 1

2. The % yield of ammonia as a function of time in the reaction:

N 2  g  3H2  g    2NH3  g , H  0
  
at (P, T1) is given below:

T1
% Yield

time
If this reaction is conducted at (P, T2), with T2 > T1, the % yield of ammonia as a function of time
is represented by

T2 T2
% Yield
% Yield

T1
T1

(A) (B)
time time

T1 T1
% Yield

% Yield

T2 T2

(C) (D)

time time

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100829)-7

3. At starting 20 moles of H2, 8 moles of I2 was taken in a 10 litre flask and at equilibrium the density
of HI was found to be 0.0384 g mL – 1 . What will be the degree of dissociation for HI keeping the
temperature and pressure constant:
(A) 45% (B) 88%
(C) 32% (D) 25%

4. Following two equilibria are established on mixing two gases A2 and C

(i) 3A 2  g   A 6  g K p = 1.6atm -2
(ii) A 2  g  + C  g   A 2C  g 
If A2 and C are mixed in 2:1 ratio (molar), at equilibrium, the total pressure is 1.4 atm and partial
pressure of A6 is 0.2 atm at equilibrium.
Which is false
(A) partial pressure of ‘A2’ at equilibrium is 0.5 atm
(B) KP for second reaction is 1.5 atm-1
(C) partial pressure of ‘C’ at equilibrium is 0.4 atm.
(D) partial pressure of ‘A2C’ at equilibrium is 0.4 atm.

5. Lanthanum (III) chloride is a white hygroscopic powder which attains the following equilibrium in a
sealed vessel:
LaCl3  s  H2O  g     LaClO  s  2HCl  g 

More water vapour is added and the equilibrium is allowed to re-establish. If at new equilibrium,
PH2O has doubled, PHCl increases:
(A) 2 times (B) 2 times
(C) 3 times (D) 3 times

6. There are three stretchable balloons are filled up to their maximum capacity

PCl5  PCl3  Cl2 N2  3H2  2NH3 CO Cl2  COCl

 g g  g  g  g  g  g  g 2
 g

(I) (II) (III)

Which of the balloon will not burst if small amount of N2 is added in these balloons?
(A) I (B) II
(C) III (D) None

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-8

SECTION – 2 : (One or More Than One Options Correct Type)

This section contains 3 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.
7. When N2O5 is heated, it dissociates as N 2 O5  N 2 O3  O2 , Kc = 4.5. At the same time, N 2O3 also
decomposes as N 2 O3  N 2 O  O2 . If initially 4 moles of N2O5 are taken in a 1 l flask and allowed
to attain equilibrium, concentration of O2 was found to be 4.5M. Which of the following is/are
correct?
5
(A) Equilibrium concentration of N2O is .
3
7
(B) Equilibrium concentration of N2O3 is .
6
(C) Equilibrium constant of the reaction N 2 O3  N 2 O  O2 is approximately 6.428.
(D) All are correct.
8. Hydrazine was taken in a constant volume container at 27 oC and 0.3 atm and equal moles of
oxygen was injected, sealed and finally heated to 1000 K where the following equilibria
established
N2H4  3O2    2NO2  2H2O K p1
N2H4    N2  2H2 Kp 2

N2H4  H2    2NH3 K p3

If the gaseous mixture at equilibrium is passed through moisture absorbent, a decrease of 360
mm in the equilibrium pressure was observed. Now if the dried gaseous mixture is passed
through ammonia absorber a further decrease of 20 mm in the equilibrium pressure was
observed. If K p1  4 , the correct statements from the following are:
(A) Equilibrium pressure of H2 is 0.4528 atm
(B) Equilibrium pressure of N2H4 IS 0.517 atm
(C) K p2  9.24  10 2 atm2
(D) K p3  2.95  10 3
9. For the dissociation:
1
N2O5  g    2NO2  g   O2  g 
2
if,
M = Molecular mass of N2O5(g)
D = Vapour density of equilibrium mixture
Po = Initial pressure of N2O5(g)
Identify the ‘true’ statements:
PoM
(A) Equilibrium pressure can be expressed as
2D
2PoD
(B) Equilibrium pressure can be expressed as
M
Po  M  2D 
(C) Degree of dissociation can be expressed as
3D
(D) Increase in temperature will lower the magnitude of D

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100829)-9

SECTION – 3: (Numerical Answer Type)

This section contains THREE 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. xxxxx.xx).

10. Ammonium carbamate is heated at 227oC in a closed vessel of volume 8210 ml containing 0.2
mole of Argon gas. The manometer attached to the vessel shows a pressure of 1 atm. Find K p for
the decomposition of ammonium carbamate.
NH2 COONH4  s     2NH3  g   CO 2  g 

11. A 2 lit flask, initially containing one mol of each CO and H 2O, was sealed and heated to 700 k,
where the following equilibrium was established.
CO(g)  H2 O(g)    CO2 (g)  H2 (g) ; KC = 9
Now the flask was connected to another flask containing some pure CO 2(g) at same temperature
and pressure, by means of a narrow tube of negligible volume. When the equilibrium was
restored, moles of CO was found to be double of its moles at first equilibrium. Determine volume
of CO2(g) flask, (In litre).

12. Consider the reaction A    B at 1000 K. At time t’, the temperature of the system was
increased to 2000 K and the system was allowed to reach equilibrium. Throughout this
experiment the partial pressure of A was maintained at 1 bar. Given below is the plot of the partial
pressure of B with time. What is the ratio of standard Gibbs energy of the reaction at 1000 K to
that at 2000 K?

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-10

PART III : MATHEMATICS

SECTION – 1: (Only One Options Correct Type)

This section contains 6 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE is correct.

m n
1. One of the roots of equation 2000x6 + 100x5 + 10x3 + x – 2 = 0 is of the form , where m is
r
non–zero integer, and r and n are relatively prime natural number, then the value of (m + n + r) is
(A) 200 (B) 205
(C) 195 (D) 190

1
2. Consider the equation x 4  2ax3  x 2  2ax  1  0 , where a  R . Also, range of f  x   x  is
x
 ,  2   2,   . If equation has at least two distinct positive real roots, then all possible values
of ‘a’ are
 1 5 
(A)  ,   (B)  ,  
 4 4 
 3
(C)  ,   (D) none of these
 4

3. Sum of all the roots of the equation x2 – 2x + |x – 1| – 5 = 0 is

(A) 0 (B) 2
(C) 1 (D) 5

2a 2  x 2 2x 1
4. The set of real values of a for which the equation    0 has a unique
3
a x 3 2
ax  a  x 2 xa
solution is
(A) (– , 1) (B) (–1, )
(C) (–1, 1) (D) R – {0}

Space for rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100829)-11

5. Roots of a quadratic equation x2 + 5x + 3 = 0 are 4cos2  + a, and 4sin2 + a. Another quadratic

equation is given as x2 + px + q = 0 where p, q  N and p, q  [1, 10]. If roots of second quadratic
equation are real then the probability that they are 4cos 4  + b and 4sin4  + b is
1 3
(A) (B)
16 32
1 5
(C) (D)
32 100

x2 a
6. If  4 x  13  sin , then a is equal to
3 x
 
(A) (2n  1) , n  Z (B) 3( 4n  1) ,nZ
2 2
(C) 3 (1 + 4n) , n  Z, n  Z (D) None of these

SECTION – 2: (Multi Correct Choice Type)

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE OR MORE may be correct.

7. Given ax2 + bx + c  0, bx2 + cx + a  a and cx2 + ax + b  0, where a  b  c and a, b, c  R.

a2  b2  c 2
Now can not take values
ab  bc  ca
2 1
(A) (B)
3 2
3 9
(C) (D)
2 2

8. If the roots of the equation x2 – px – 1 = 0 and x2 – qx – 1 = 0 form (in a suitable order) an

arithmetic progression with four members then
2 2 4 4
(A) p  ,q (B) p  ,q
3 3 3 3
4 4 2 2
(C) p  ,q (D) p  ,q
3 3 3 3

Space for rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100829)-12

9. If the product of the roots of the equation 2x2 + ax+4 sina = 0 is 1, then roots will be imaginary if
 7  
(A) a  R (B) a   , 
 6 6
  5 
(C) a   ,  (D) None of these
6 6 

SECTION – 3: (Numerical Answer Type)

This section contains THREE 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. xxxxx.xx).

10. If all the roots of the equation x1008  a1x1007  a2 x1006  .....  a1006 x 2  1008x  1  0 are real, then
a2
is
a1

11. The real numbers ,  satisfy the equation 3 – 32 + 5 – 17 = 0; 3 – 32 + 5 + 11 = 0.

Then  +  is equal to

12. The value of ‘n’ for which the equation 9x4 – 12x3 + nx2 – 8x + 4 is a perfect square

Space for rough work

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