08-06-2025

2001CJA101031250002 JA

PHYSICS

SECTION-I(i)

1) A pendulum bob of weight 1N is held at an angle θ from the vertical by a 1N horizontal force F as
shown. The tension in the string supporting the pendulum bob (in newtons) is

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

2) Work done when a force acting on a particle takes it from the point
to the point is

(A) –3 J
(B) –1 J
(C) zero
(D) 2 J

3) Two point charges + 9e and +e are kept 16 cm. apart from each other. Where should a third
charge q be placed between them so that the system remains in the equilibrium state :–

(A) 24 cm from + 9e
(B) 12 cm from + 9e
(C) 24 cm from + e
(D) 12 cm from + e

4) A particle starts from rest. What is the displacement of particle in 4th second, if its acceleration is
2m/s2 :-

(A) 16 m
(B) 7 m
(C) 9 m
(D) 8 m

SECTION-I(ii)
1) A particle moves with constant speed v along a regular hexagon ABCDEF in the same order .
Then the magnitude of the average velocity for its motion from A to :

F is
(A)

D is
(B)

(C)
C is
(D) B is v .

2) A projectile is projected at an angle of elevation α. After t seconds, it appears to have an angle of

elevation β as seen from the point of projection. Which of the following are correct?

(A) the y-coordinate of the position of the projectile at time t is , where v is the
velocity of projection
the x-coordinate of the position of projectile at time t is , where v is the velocity of
(B)
projection

(C)
tan β = tan α – , where v is the velocity of projection

(D)
velocity of projection is

3) A trolley of mass 8 kg is standing on a frictionless surface inside which an object of mass 2 kg is

suspended. A constant force F starts acting on the trolley as a result of which the string stood at an

angle of 370 from the vertical. Then :

(A) acceleration of the trolley is 40/3 m/sec2

(B) force applied in 60 N
(C) force applied is 75 N
(D) tension in the string is 25 N

4) In the system shown in the figure m1 > m2 . System is held at rest by thread BC. Just after the
thread BC is burnt (neglect friction) :

(A) acceleration of m2 will be upwards

(B)
magnitude of acceleration of both blocks will be equal to g
(C) acceleration of m1 will be equal to zero
(D) magnitude of acceleration of two blocks will be non-zero and unequal.

5) The two blocks A and B of equal mass are initially in contact when released from rest on the
inclined plane. The coefficients of friction between the inclined plane A and B are μ1 and μ2

respectively.

(A) If μ1 > μ2, the blocks will always remain in contact

(B) If μ1 < μ2, the blocks will slide down with different accelerations (if blocks slide)
(C) If μ1 > μ2, the blocks will have a common acceleration (μ1+ μ2) g sin θ

(D)
If μ1 < μ2, the blocks will have a common acceleration sin θ

6) Work done by force of friction on the object :

(A) can be zero

(B) can be positive
(C) can be negative
(D) information insufficient

SECTION-II(i)

Common Content for Question No. 1 to 2

Stem Questions: Two blocks A and B are placed on frictionless surface as shown in figure

mA = 4 kg
mB = 8kg

1)

Acceleration of blocks is ______ m/s2

2) Tension in string

Common Content for Question No. 3 to 4

A 2 kg block is pressed against a rough wall by a force F ah shown in figure. (Take g = 10 m/s2)

3) If F = 30 N, find friction on block.

4) If F = 20 N, find friction on block.

Common Content for Question No. 5 to 6

A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction

between A & B is 0.5. (t is time in seconds. Initial velocities are zero)

5) The Block A shart sliding on B at T = _______ sec

6) The acceleration of block A at t = 5 sec is _______

SECTION-II(ii)

1) A ball is projected at an angle of 60° to the horizontal. If ball has a velocity of 50 m/s at the
highest point then ball was projected with velocity v (in m/s), find the value of .

2) A particle moves such that x (m) = t2 + t – 3 ; where t is in second. Find its average velocity(in
m/s)from t = 2 sec to t = 5 sec.

3) A block of mass 2 kg is in equilibrium under action of three forces. If

the force of 8N is removed and other remain unchanged then what will be magnitude of acceleration
of block in m/s2.

CHEMISTRY

SECTION-I(i)

1) How many moles of Na+ ions are in 20 mL of 0.40 M Na3PO4 :

(A) 0.0080
(B) 0.024
(C) 0.050
(D) 0.20

2) Out of Molarity (M), Molality (m), % w/v and Mole fraction (x), those independent of temperature
are:

(A) M,m
(B) M, % w/v
(C) m,x
(D) M,x

3) Correct order for stability of intermediate is :

(i) (ii) (iii) (iv)

(A) i > ii > iii > iv
(B) ii > iii > iv > i
(C) i > iii > iv > ii
(D) i > iv > iii > ii

4) Select the most basic compound in aqueous medium out of following :

(A) NH3
(B) MeNH2
(C) Me2NH
(D) Me3N

SECTION-I(ii)

1) The incorrect statement(s) regarding 2M MgCl2 aqueous solution is/are (dsolution = 1.09 gm/ml)

(A) Molality of Cl¯ is 4.44 m

(B) Mole fraction of MgCl2 is exactly 0.035
(C) The conc. of MgCl2 is 19% w/v
4
(D) The conc. of MgCl2 is 19 × 10 ppm

2) Which of the following have same mass -

(A) 0.1 mole of O2 gas

0.1 mole of SO2 gas
(B)
22
(C) 6.023 × 10 molecules of SO2 gas
23
(D) 1.204 × 10 molecules of O2 gas

3) Dichromate ion in acidic medium oxidizes stannous ion as:

(A) The value of x : y is 1 : 3

(B) The value of x + y + z is 18
(C) a : b is 3 : 2
(D) The value of z – c is 7

4) Correct statement(s) is/are :

(A) Resonating structures are real & resonance hybrid is hypothetical
(B) Inductive effect is distance dependent effect
–
(C) Acetate ion (CH3–COO ) contain identical resonating structures
(D) Resonance involve permanent displacement of π electron density due to σ-bond

5) Which carbocation(s) is/are more stable than :

(A)

(B)

(C)

(D)

6) Among the following resonating structures, in which option(s), (I) is more stable than (II) ?

(A)

(B)

(C)

(D)

SECTION-II(i)

Common Content for Question No. 1 to 2

Aruna was studing the interaction between various chemicals. She found that when she mixed
potassium bromate (KBrO3) with potassium bromide, (KBr) and acidified the solution with H2SO4,
vapours of Br2 were evolved. On addition of KΙ to the reaction mixture, deep colour was obtained due
to formation of iodine, and this could be titrated with sodium thiosulphate (Na2S2O3) using starch
indicator
1) The equivalent weight of Br2 in the comproportionation reaction is [Atomic weight of Br = 80]

2) Starting with 10 ml of 0.01M KBrO3 and using excess of KBr, H2SO4 and KI, the volume of 0.05M
Na2S2O3 required will be -

Common Content for Question No. 3 to 4

the compound in which cyclic delocalisation of Huckel's (4n + 2) π-electrons is present, is classified
into aromatic compound on the other hand compound which has cyclic conjugation of 4n π-electrons
is classified into anti aromatic compound.

3) How many compound are aromatic compound?

(1) (2) (3) (4)

(5) (6) (7)

4) How many compound are anti-aromatic compound?

(1) (2) (3) (4)

(5) (6) (7)

Common Content for Question No. 5 to 6

Concentration terms are the different ways used to express the amount of solute present in a
solution or mixture. These terms provide quantitative descriptions of how much of a substance is
dissolved or mixed with another.

5) What is mole fraction of N2(g) in a mixture where N2(g) and O2(g) are 20:80 by volume
respectively.

6) A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the
KOH solution.
(Molar mass of KOH = 56 g mol–1)

SECTION-II(ii)

1) The number of moles of neutrons found in 2 gms of Oxygen atoms will be:

2)

Total number of groups which can show –I effect when attached to benzene ring :

(a) –OH (b) –NH2

(c) (d) –Cl

(e) (f) –O–

(g) –CH3 (h)

(i) (j) –C≡N

(k)

3) How many resonating structures of carbocation 'M' as shown below are possible (excluding given

resonating structure) :

MATHEMATICS

SECTION-I(i)

1) If k = tan 20° + tan 40° + tan 20° tan 40°, then k2 is equal to

(A) 3
(B) 4
(C) 6
(D) 7

2) If the equations k (6x2 + 3) + rx + 2x2 – 1 = 0 and 6k (2x2 + 1) + px + 4x2 – 2 = 0 have both roots
common, then the value of (2r – p) is

(A) 0
(B) 1/2
(C) 1
(D) none of these

3) The domain of f(x) = loge |logex| is-

(A) (0, ∞)
(B) (1, ∞)
(C) (0, 1) ∪ (1, ∞)
(D) (–∞, 1)

4) The domain of the function f defined by is defined to

(A) (–∞, –1) ∪ (1, 4]

(B) (–∞,–1] ∪ (1, 4]
(C) (–∞, –1) ∪ [1, 4]
(D) (–∞, –1) ∪ [1, 4)

SECTION-I(ii)

1) The equation has -

(A) 2 solutions in (0, π)

(B) 4 solutions in (0, 2π)
(C) 2 solutions in (–π, π)
(D) 4 solutions in (–π, π)

2) The value of at , is greater than

(A) –1
(B) 0
(C) 1

(D)

3) Which of the following is/are true ?

(A)

(B) sin50° – cos40° = 0

(C) sin2 – cos2 < 0
(D) sin2 – cos2 > 0

4) Graph of f(x) = ax2 + bx + c, a ≠ 0, a, b, c ∈ R is given adjacently. Which of the following is

correct ?

(A) a(a + b + c) < 0

(B) b(b2 – 4ac) (4a – 2b + c) < 0
(C) c < 0
(D) a + 2b + 4c < 0

5) If f(x) = 3sinx + 4cosx + 5 then

(A) Maximum value of f(x) is 5

(B) Maximum value of f(x) is 10
(C) Minimum value of f(x) is –5
(D) Minimum value of f(x) is 0

6) If equation ax2 + bx + c = 0 have roots , then

(A)
cx2 + bx + a = 0 have roots
(B) ax2 – bx + c = 0 have roots

(C)
cx2 – bx + a = 0 have roots
(D) none of these

SECTION-II(i)

Common Content for Question No. 1 to 2

Let the minimum value of |x + 1| + |x – 2| is α and the greatest value of x satisfying the equation |x –

2| = is β.

1) The value of α is ____________.

2) The value of β is ____________.

Common Content for Question No. 3 to 4

Let and If and

3) The value of α is _____________.

4) The value of β is __________.

Common Content for Question No. 5 to 6

Consider the inequality . Let the smallest and largest positive integer which
satisfies the above inequality is α & β respectively.

5) The value of α is __________.

6) The value of β is ________.

SECTION-II(ii)

1) The number of integral values of 'k' for which the equation 3sinx + 4 cosx = k + 1 has a solution,
k ∈ R is

2) The minimum value of the expression 4x2 + 2x + 1 is -

3) If one of the roots of the equation x2 – bx + c =0, b, c is then bc equals

 ANSWER KEYS

PHYSICS

SECTION-I(i)

Q. 1 2 3 4
A. B B B B

SECTION-I(ii)

Q. 5 6 7 8 9 10
A. A,C,D A,B,C,D C,D A,C A,B A,B,C

SECTION-II(i)

Q. 11 12 13 14 15 16
A. 1.33 37.33 20 12 5 5

SECTION-II(ii)

Q. 17 18 19
A. 5 8 4

CHEMISTRY

SECTION-I(i)

Q. 20 21 22 23
A. B C C C

SECTION-I(ii)

Q. 24 25 26 27 28 29
A. B,D B,C,D B,C,D B,C A,B,D A,C

SECTION-II(i)

Q. 30 31 32 33 34 35
A. 96.00 12.00 3.00 2.00 0.20 1.27 or 1.29

SECTION-II(ii)

Q. 36 37 38
A. 1 8 5

MATHEMATICS
 SECTION-I(i)

Q. 39 40 41 42
A. A A C A

SECTION-I(ii)

Q. 43 44 45 46 47 48
A. B,D A,B A,B,D A,B,C,D B,D A,B

SECTION-II(i)

Q. 49 50 51 52 53 54
A. 3.00 4.50 6.50 1.00 2.00 6.00

SECTION-II(ii)

Q. 55 56 57
A. 11 75 4
 SOLUTIONS

PHYSICS

1)

2)

3)

For equilibrium
Fnet = 0
F1 = F2

4)
= 7m

5)
A to B → V

A to C →

time from A to F →

Average velocity =

6) y-coordinate is

y = v sin α t [Therefore A is correct]

x –coordinate is
x = v cos α t [Therefore B is correct]
Equation of trajectory is

⇒ [C is correct]

Rearrenging

⇒ [∴ D is correct]

7) atrolly= a0 = ...(1)
from figure we have
T cos 37° = 2g ...(2)
⇒ T = 25 N
T sin 37° = 2a ...(3)
⇒ a0 = 7.5 m/s2
from (i) we have
so, F = 75 N

8)

Initially kx0 = m1g

and, m2g + T = kx0 or m1g
when thread is burnt then we have
T=0 but x0 will remain as it is at that moment
so, acclerate block a having mass m, will be zero and accelerator of block B having mass m2 is

a= actually upwards.

9)

a = g sin θ – μg cos θ
(A) If μ1 > μ2
a1 < a2
block will always remains in contact
(B) μ1 < μ2
a1 > a2
block will move with different acceleration.
10) Work done by friction on an object can be zero, positive or negative example of positive
work done could by working of cycle.

11) 8g sin37° – 4g sin53° = 12a

48 – 32 = 12a
16 = 12a

12)

T – 4g sin53° = 4a

13)

= 24 N fk = 18 N

14) fk =0.6 × 20
= 16 N fk = 12
 15)
fmax = µmg = 0.5 × 5 × 10 = 25 N
F = ma
25 = 5a
a = 5 m/s2
F – 25 = 5a
10t – 25 = 25
10t = 50
t = 5 sec

16) F = 10 × 10 = 100 N

F = ma
25 = 5 × a
a = 5 m/s2

17) vtop is only horizontal ⇒ vcos 60° = 50 (given)

∴ velocity of projection (v)

18)

average velocity =

19)

CHEMISTRY

20) Level : Medium

Concept : Molarity =
Solution :

1 mole Na3Po4 contains 3 mole Na+

no. of moles of Na3Po4 = Molarity volume (in L)

=
 =
+
no. of moles of Na Ion =
=
Hence, correct answer is option (B)

21) Concentration terms independent of volume are independent of temperature.

22)

Correct order for stability of intermediate is :

(i) (ii) (iii)

(iv)
(+M) (–M) (+H)
(+I)

Order : (i) > (iii) > (iv) > (ii)

23)

Me2NH

24)

(A) 2M of MgCl2 = 190 gm in 1090 gm solution or 900 gm solvent

So, molality of MgCl2 =

So, molality of Cl– = 2 × 2.21 = 4.44

(B) Mole of H2O =

Mole fraction of MgCl2 =

(C) Conc. in (w/v) =

(D) conc. of MgCl2 in ppm =

25)
mass of 0.1 mol O2 will be = 0.1 × 32 = 3.2 gm
mass of 0.1 mol 5O2 will be = 0.1 × 64 = 6.4 gm
mass of 6.023 × 1022 molelcules of SO2 gas will be = 0.1 × 64 = 6.4 gm
mass of 1.204 × 1023 molelcules of O2 gas will be = 0.2 × 32 = 6.4 gm

26)

27)

Inductive effect is distance dependent effect

Resonance involve permanent displacement of π electron density due to σ-bond

28)

29)

30)

BrO3– + Br– → Br2

Anode (Br– )5

Cathode : 6H+ + BrO3– + 5e– Br2 + 3H2O

______________________________________________
5Br– + BrO3– + 6H+ 3Br2 + 3H2O
______________________________________________
3 moles 5e–
1 mole e–

nF =

Equivalent weight = =
= 32 × 3 = 96

31) 5Br– + BrO3– + 6H+ → 3Br2 + 3H2O

Br2 + 2I– I2 + 2Br–
I2 + 2Na2S2O3 Na2S4O6 + 2NaI

0.05

Vml =

32) (3), (5), (7)

33) (1), (4)

34)

35)

6.90 M solution of KOH contains 6.90 moles of KOH is 1000 ml. of solution.

Wt. of KOH in solution = 6.90 × 56 = 386.4g

Wt. of KOH in 1000 ml. solution = 386.4g

Since the solution is 30% by weight, it means that 30g of KOH are

present in 100 g of solution. 386.4g of KOH is present in =

= 1288g of solution

Density =

36)

= Moles of O atom.

Moles of neutrons in O atoms =

37) Explanation:
To count no. of groups showing –I.E
Concept:
Inductive effect.
Solution:
–I.E showing group are e– withdrawing groups having more electro negative atom attached to
CSP3 Carbon
a, b, c, d, e, i, j, k, are –I groups.
 Final Answer: 8

38)

MATHEMATICS

39)

tan 60° = tan(40° + 20°) =

tan 20° + tan 40° + tan 20° tan 40°
=k
⇒ k2 = 3

40)

(6k + 2)x2 + rx + 3k – 1 = 0
6x2 + px – 1 = 0

2r – p = 0

41)

|logex| > 0 ∩ x > 0

x ∈ R – {1}
x ∈ (0, ∞) – {1}

42)

Given that :
f(x) is defined if
4 – x ≥ 0 or x2 – 1 > 0 ⇒ –x ≥ – 4 or (x – 1)(x + 1) > 0
⇒ x ≤ 4 or x < –1 and x > 1
∴ Domain of f(x) is (–∞, –1) ∪ (1, 4]

43)
44)

45)

(B) sin(90° – 40°) – cos40°

cos40° – cos40° = 0
(C) and (D)

45° < θ < 90°

sinθ > cosθ

46)

(i) a > 0 (ii) c < 0 (iii) D > 0 (iv) b < 0

(v) f(1) < 0
a+b+c<0

(vi)

a + 2b + 4c < 0

47)

Range = [–5, 5] + 5
= [0, 10]

48)

(A) Replace
(B) Replace x → –x

49)

Put x = 2 or x = –1
Minimum value α = 3

Greatest Value

50)

Put x = 2 or x = –1
Minimum value α = 3

Greatest Value

51)

a = 25
b = log7 40
c = log740

52)

a = 25
b = log7 40
c = log740

53)

x ∈ (–2, 0) ∪ (1, 2] ∪ {6}

α = 2, β = 6

54)

x ∈ (–2, 0) ∪ (1, 2] ∪ {6}

α = 2, β = 6

55)

3 sinx + 4 cosx = k + 1
⇒ k + 1 ∈ [– 5, 5]
⇒ k ∈ [–6, 4]
No. of integral values of k = 11

56)
minimum value =

57) ,
α + β = 4 and αβ = 1
bc = 4