(Batches: e-SANKALP-2325 S1, T1 & P1)

IIT – JEE, 2325 Paper Code

100850
(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 & P1-XI-PCM-(100850)-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 uniform straight rod of mass M is free to rotate about a vertical axis

through one of its end perpendicular to its length. It is given an angular
velocity 0 about the vertical axes. And insect of mass M/3 starts moving
along the rod away from the axes of rotation with constant speed L0
relative to rod, where L is the length of the rod. By the time insect has
reached to the other end, rod has turned through an angle equal to
(A) /4 (B) /2
(C) 3/4 (D) 

2. A smooth horizontal rod of mass 6m and length L has a sleeve of

mass 4m free to slide over it placed at the middle as shown. The rod 4m
is free to rotate about a vertical axis through one end. Initial angular
speed 0 is provided to rod by sharp impulse. Find the speed of the
sleeve just at the instant it gets free from the rod 6m
L/2
(A) 0L/2 (B) 0L
3 5
(C) 0L (D) (0L)
8 8

3. A smooth cylindrical glass lies on the ground. A small particle of mass R

m lies on the bottom surface touching circumference as shown in the

figure. Find the minimum value of v 0 (along the circumference) such
that particle just reaches at the top of glass. h
2
(A) 2gh (B) 2gh
3
2gh r
(C) gh (D) 2
3 v0 r = R/2

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4. An ideal inextensible string is wrapped over the disc of mass m and radius
R. The other end of the string is connected to mass m. The string is passing A
over an ideal pulley A as shown in the figure. At any time t, mass m and
disc are moving downward with acceleration of magnitudes a1 and a2
respectively. The disc is rotating clockwise with angular acceleration of m 
m
magnitude . There is no slipping between string and disc. Choose the R
INCORRECT option. a1 a2

(A) a1 = a2 (B) R > a1

(C) R > a2 (D) R < a2

5. A rod and a block are of same mass. Initially rod is in horizontal

position. When system is released from this position. The
acceleration of tip of the rod is m
3g 3g
(A) (B)
2 4
3g 3g m
(C) (D)
8 11

6. A uniform rod of mass M is hinged about hinge O and y

its other end is tied with a string. The string is
horizontal and whole system is in equilibrium. The rod
makes an angle of 45 with horizontal. Choose the
correct option.

O 45o
x
The rod is uniform and has
mass M.
(A) The horizontal and the vertical component of the hinge reaction are equal to NX = Mg/2;
NY = Mg/2
(B) The horizontal and the vertical component of the hinge reaction are equal to NX = Mg;
NY = Mg
(C) If the string is cut, then the value of NX and NY immediately after cutting the string is equal to
NX = 3Mg/8 ; NY = 5Mg/8
(D) If the string is cut, then the value of NX and NY immediately after cutting the string is equal to
NX = 3Mg/8; NY = Mg

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e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-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.

1
7. A uniform semicircular wire of mass M = kg and radius R = 1m is free to rotate about a fixed
2
horizontal axis coinciding with the diameter passing through open ends. First the wire is taken
aside such that its plane becomes horizontal and then it is released from rest. Choose the correct
option(s), when vertical component of velocity of the centre of mass is maximum (take g = 10
m/s2, 2 = 10)
(A) Angular displacement of wire is cot 1 2
(B) Vertical component of force exerted by the axis on the wire is 5 2 N
(C) Vertical component of force exerted by the axis on the wire is 10 2 N
(D) Horizontal component of the force exerted by the axis on the wire is 8N

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 e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-5

8. A uniform rod AB of mass M is attached to a hinge at one end A, and released from rest from the
horizontal position. The rod rotates about A, and when it reaches the vertical position the rod
strikes a sphere of mass m and radius r initially at rest on the smooth horizontal surface as shown
in the adjacent figure. The impact is along the horizontal direction and perfectly elastic.
If at the moment of impact the lowest end of the rod is very close to the smooth horizontal
surface. After the impact, the sphere moves along the horizontal and the rod, subsequently rises
to a maximum of 60 with the vertical. Choose the correct statement(s) from the following, taking
6 2
into account the information given above. The length of the rod equals 2r . (r = m)
10
Initial final
position A position
m A m

60
r r
B

B
(A) The ratio M/m is 3/2
(B) The ratio m/M is 2/3
(C) The speed of the sphere just after collision is 6 m/s
(D) The speed of the sphere just after collision is 3 m/s

9. A solid sphere of mass m is at rest on a smooth m

horizontal surface and there is a very small cylindrical v0
60
hole in the solid sphere upto centre of the solid sphere. Solid
A small particle of mass m enters in the hole with sphere
velocity v0 and sticks with the sphere at the centre of
solid sphere. Size of hole is such small that a point mass P Smooth Q Rough
can just move inside it. (collision is such that sphere
does not lose contact with the surface after collision).
Surface on the left side of point Q is smooth and right
side is rough as shown in the figure. Then (Assume
moment of inertial of sphere Icm = 2mR2/5)
(A) linear momentum conserved in horizontal direction.
(B) angular momentum conserved about point P which is point of ground.
5 v0
(C) after collision if pure rolling starts then angular velocity of solid sphere is .
24 R
(D) just after collision velocity centre of mass of the system is v 0/4

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e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-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. A thread is passing through a hole at the centre of a frictionless table. At the upper end a block of
mass 0.5 kg is tied and a block of mass 8 kg is ties at the lower end which freely hanging. The
smaller mass is rotated on the table with a constant angular velocity about the axis passing
through the hole so as to balance the heavier mass. If the mass of the hanging block is changed
K
from 8 kg to 1 kg. The fractional change in the angular velocity of the smaller mass is  , so
4
that it balances the hanging mass again. Find the value of K.

L M, L/3
11. A smooth disc of mass M and radius is placed at rest horizontally
3 M
on a smooth horizontal surface. A massless pin is fixed at point P at a
O
distance L/2 from centre O of the disc as shown in the figure. Now a
0 L/2
thin uniform rod of mass M and length L is placed horizontally on the
surface of the disc parallel to the line OP such that its mid point and P
centre O of the disc just coincide as shown in figure. Now rod has
given angular velocity 0 = 24 rad/sec in counter clockwise direction
as shown. As a result, the end of the rod strikes the pin P and sticks to
it rigidly. Calculate the angular velocity of disc just after collision.

12. A uniform square plate of mass m = 100 gm and side a = 24 cm can freely rotate about a vertical
axis passing through one edge. It is initially at rest. A particle of mass m = 100 gm is moving
horizontally and perpendicular to the plane of the plate with velocity u = 70 cm/s. The particle
collides with the plate elastically at the centre of the plate. Find the angular velocity (in rad/s) of
the plate just after collision.

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 e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-7

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. In what volume should 0.03 M HCl (V1) and 0.03 M of a weak base BOH (V2) be mixed to make
200 ml of a buffer at pH = 7? (Kb of BOH = 9.8 × 10 – 8 at 25oC) log9.8 = 0.99 and antilog (– 0.01)
= 0.977.
(A) V1 = 4.56 ml, V2 = 195.44 ml (B) V1 = 100 ml, V2 = 100 ml
(C) V1 = 180.5 ml, V2 = 19.5 ml (D) V1 = 64.3 ml, V2 = 135.7 ml

2. The electrolytic oxidation of N2O4 was carried out in a solution buffered by NaHCO3 and Na2CO3
(dissolved in equal proportion by mols). The reaction involved was:
N2 O4  2H2 O   2NO3  4H  2e 
100 ml of 1M solution of N2O4 buffered initially at pOH = 3, was oxidized, with reaction above
going to completion. The total anionic concentration initially, i.e. HCO3- + CO2- 
3  was 4 M. The pH

of the solution after oxidation will be: (For H2CO3. K a1  10 6 and K a2  10 11 ) :
(A) 5.7 (B) 6.0
(C) 6.3 (D) 11.0

3. 100 ml of 0.5 M hydrazoic acid (HN3, Ka = 3.6 x 10–4) and 400 ml of 0.1 M cyanic acid (HOCN),
Ka = 8 x 10–4) are mixed, which of the following is false for the final solution?
(A) [H+] = 10–2 M (B) [N3–] = 3.6 x 10–3 M
– –3 +
(C) [OCN ] = 6.4 x 10 M (D) [H ] = 1.4 x 10–2 M
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4. A 1.025 gm sample containing a weak acid HX (mol. wt = 82) is dissolved in 60 ml H2O and titrated
with 0.25 M NaOH solution. When half of the acid is neutralized the pH was found to be 5.0 and the pH
at equivalence point is 9.0. Calculate the percentage of HX in the original sample. (Assume law of
conservation of volume is obeyed)
(A) 40% (B) 80%
(C) 65% (D) 78%

5. 3 g of CH3COOH is dissolved in 100 mL water to prepare a solution X. Now, 2 gm of NaOH is added

to the solution X to form solution Y. Finally, 2.45 g of H2SO4 is added into Y to prepare solution Z,
Ka (CH3COOH) = 1.8 × 10–5. [log 3 = 0.48]
What is the pH of solution (Z)?
(A) 5.52 (B) 6.45
(C) 3.15 (D) 2.52
o
6. The equilibrium constant for the following reaction at 25 C
CN  HAc   HCN  Ac  is x  10y. The value of x + y is ….?

–5 3 –9
[Given that Ka(HAc) = 1.8 × 10 mol/dm ; Ka (HCN) = 0.45 × 10 ]
(A) 4 (B) 2.5
(C) 8 (D) 4.5
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 e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-9

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. The pH of mixture obtained by mixing V ml each of 0.1 M salicylic acid and 0.1 M NaOH is x. The
Ka1, and Ka2 of salicylic acid are 10 – 4 and 10 – 7 respectively at T Kelvin [log 3 = 0.48]
(A) The value of x is 5.5
(B) The value of x is 1.5
(C) The pH of above mixture is independent of concentration of salt formed
(D) If V mL 0.1 M NaOH is further added to the above mixture pH of the resultant solution
becomes 10.

8. Identify the correct statements w.r.t. the titration curves shown below:
14

pH

Weak acid

Strong acid

Volume of strong base added

(A) equivalence point with strong acid is at pH = 7

(B) equivalence point with weak acid is at pH = 7
(C) phenolphthalein (8 – 9.8) can be used as indicator in both titrations
(D) chlorophenol red (4.8 – 6.4) can be used as indicator in both titrations.

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9. Following titration method is taken to compute stepwise ionisation constant of a weak dibasic acid ‘X’

O OH
C

' X' 

OH

25 ml of a dilute aqueous solution of ‘X’ is titrated with 0.2 M NaOH(aq) and pH measured

Step Volumes of NaOH added pH

I 8.12 ml 4.57
II 16.24 ml 7.02 (at first

equivalence point)

Which of the following is/are correct?

(A) First step reaction can be:

COOH COOH

+
+ H2O + H3O .

OH O
(B)  
pK a1   logK a1 of ‘X’ is 4.57.

(C)  
pKa2   log K a2 of ‘X’ is approx 9.5.

(D) pH at half of the second equivalence point (after first equivalence point) will be 9.47.

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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. 100 ml, 0.1 M Na3PO4 solution is added in sequence [first Na3PO4 solution is mixed with (i) and
the resulting mixture with (ii) and resulting mixture finally with (iii)] into the following reagents:
( K a ,K a and K a of H3PO 4 are 10-4, 10-7 and 10-11) respectively.
1 2 3

(i) 100 ml, 0.1 M NaH2PO4 solution

(ii) 100 ml, 0.05 M Ca(OH)2 solution
(iii) 100 ml, 0.1 M H2SO4 solution
x+y
If pH, after step I and step III are x and y, respectively then what is the value of ?
2

11. What is the pH when 200 mL solution of 10- 6 M acetic acid is titrated with a solution of 89 mL
solution of 10-4 M NaOH. (log3 = 0.48)

12. For a triprotic acid H3A, K a1 is 10 – 2 , K a2 is 10 – 5 and K a3 is 10 – 11. At what pH HA2 – will be at its
maximum concentration?
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e-Sankalp2325 S1, T1 & P1-XI-PCM-(100850)-12

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.

1. a, b, c are positive integers forming an increasing G.P and b – a is a perfect cube and log6a +
log6b + log6c = 6, then a + b + c =
(A) 100 (B) 111
(C) 122 (D) 189

2
2. The sequence a1, a2, a3, … satisfies a1 = 1, a2 = 2 and an + 2 =  an , n = 1, 2, 3, … the value
an1
22009 n!
of  a2012 is (where n Cr  )
2011 r ! n  r !
2010 2011
(A) C1005 (B) C1006
2011 2012
(C) C1005 (D) C1006

1 1 1 1 1 1 1 1
3. Given two series S1 = 1     ... and S2 =    ... , then
2 3 4 1001 501 502 503 1001
(A) S1 > S2 (B) S1 < S2
(C) S1 = S2 (D) S1 + S2 = 501.

4. A parabola touches two given straight lines originating from a given point. The locus of mid–point
of portion of any tangent. Which is intercepted between the given straight line is
(A) parabola (B) ellipse
(C) straight line (D) circle

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5. The circum circle of ABC is x2 + y2 – 5x – 4y + 6 = 0, if a parabola (k + 1)y2 = x have sides AB,

BC, CA as tangents then the value of k is/are
7 11
(A) – (B) –
8 13
2 4
(C) (D)
3 7

6. Let x + y – 4 = 0, x – y – 2 = 0 and x = 2 are the tangent of a parabola then the locus of focus of
the parabola is
(A) x2 + y2 – 4x + 2y – 7 = 0 (B) x2 + y2 – 4x – 3y + 5 = 0
2 2
(C) x + y – 4x – 2y + 4 = 0 (D) x2 + y2 + 4x + 3y + 5 = 0

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.

a  bk b  ck c  dk
7. If   where k  0; a, b, c, d > 0  1, then
a  bk b  ck c  dk
(A) d, a, c, b are in A.P. (B) b, a, d, c, are in H.P.
(C) logae, logbe, logce, logde are in H.P. (D) a, b, c, d are in G.P.

8. Consider a parabola (y – )2 = 4(x – ) ,    R+. P is a point on parabola such that normal at
G (on parabola) passes through P, Q is the middle point of SG and X is the foot of the directrix
[Directrix and Axis meet in X; A is vertex, ‘S’ is focus], then QX2 – QP2 = (AS)2 where  lies in
(A) (0, 4) (B) (2, 5)
7 9
(C) (3, 7) (D)  , 
 2 2

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9. Tangent drawn at point P(1, 3) of a parabola intersects its tangent at vertex at V(–1, 5). If R(–5, 5)
is a point on SP; where S is focus of the parabola then
1
(A) slope of directrix is (B) radius of circumcircle of SVP is 10 units
3
(C) tangent cuts the axis of parabola at (–3, 7) (D) focus is (–2, 4)

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. The longest geometric progression that can be obtained from the set (100, 101, …., 1000) has
the number of terms equal to ________
2
11. The number of chords of y = 16(x + 4) which subtends right angle at (0, 0) and touches the
x2 + y2 = 1 is/are _____

10  
3r r! 3r 2  5r  1 3m  n!  12
12. If  r 2  3r  2
is equal to
12
, then |m – n| is equal to _____
r 0

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