27-07-2025

2401CJA101001250016 JA

PART-1 : PHYSICS

SECTION - I (i)

1) Two masses are connected by a string which passes over a pulley accelerating upward at a rate 5
m/s2 as shown. If aceleration of mass 1 is a1 = 2 m/s2 upwards then the acceleration of mass 2 is :-

(A) 3 m/s2 upwards

(B) 3 m/s2 downwards
(C) 8 m/s2 upwards
(D) 8 m/s2 downwards

2) Two persons A and B are sitting in two cars 1 and 2 such that the velocity of the cars is given by

and respectively. The pseudo force on B as observed by A at t = 1

sec will be : (mA = 70 kg, mB = 50kg)

(A)

(B)

(C)

(D)

3) If velocity of block B in the given figure is 300mm/s towards right. Then velocity of A will be :-

(A) 200 mm/s

(B) 100 mm/s
(C) 450 mm/s
(D) 150 mm/s

4) An object starts moving from rest on a circle of radius 2m, with constant tangential acceleration
2m/s2. The time after which acceleration of object makes angle 45° from tangent is

(A) 1 sec
(B) 2 sec
(C)

(D)

SECTION - I (ii)

1) In the system shown in the figure m1 > m2 . System is held at rest by thread BC . Just after the

thread BC is burnt :

(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.

2) 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 37 from the vertical (bob at rest relative to trolley) Then :

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

(B) force applied is 60 N
(C) force applied is 75 N
(D) tension in the string is 25 N
3) A spring connects by two particles of masses m1 and m2. A horizontal force F acts on m1. Ignoring

friction, when the elongation of the spring is x then:

(A)

(B)

(C) F = m1a1 + m2a2

(D) None of these

SECTION - I (iii)

Common Content for Question No. 1 to 2

Comprehension-4
A point moves in plane X-Y according to law x = a sin ωt and y = (a – a cos ωt) where a is a positive
constant and t is the time.

1) The magnitude of velocity of body at any instant of time t is:

(A) aω cos ωt
(B) aω
(C) aω sin ωt
(D) None

2) The magnitude of acceleration of the body at any instant is :

(A) aω2
(B) aω

(C)

(D)

Common Content for Question No. 3 to 4

Two smooth blocks are placed at a smooth corner as shown. Both the blocks are having mass m. We
apply a force F on the block A. Block A presses the block B in the normal direction, due to which
pressing force on vertical wall will increase, and pressing force on the horizontal wall decrease, as
we increase F. As soon as the pressing force on the horizontal by block B becomes zero, it will loose
the contact with the ground. If the value of F is further increased, the block B will accelerate in
upward direction and simultaneously the block A will move toward right (θ = 37° with horizontal).
3) If acceleration of block A is 'a' rightward, then acceleration of block B will be :

(A)
upwards

(B)
upwards

(C)
upwards

(D)
upwards

4) What is minimum value of F, to lift block B from ground ?

(A)
mg

(B)
mg

(C)
mg

(D)
mg

SECTION - II

1) With what minimum acceleration (in m/s2) mass M must be moved on frictionless surface so that
m remains at rest with respect to 'M' as shown. The coefficient of friction between M and m is 0.4. (g

= 10 m/s2)

2) A 5kg block has a uniform rope of mass 2kg attached to its underside and a 3kg block is
suspended from the other end of the rope. An external force of 80N is applied on the 5kg block in
upward direction as shown in figure. If T is tension at mid point of rope and whole system is in mid
air above the surface. Find the value of T (in N).

3) A block of mass 2 kg is resting over another block of mass 6 kg. 2 kg block is connected to one
end of a string fixed to a vertical wall as shown. If the coefficient of friction between the block is 0.4,
the force (in N) required to pull out the 6 kg block with an acceleration of 1.5 m/s2 is ________ (g = 10

ms–2)

4) Two blocks of masses 2 kg and 3 kg are placed in a lift which moves up with an upward
acceleration of a = 2 m/s2 as shown in figure. If an external force of 15 N is acting on 2 kg block, find

the normal reaction between blocks in newton. [g = 10 m/s2]

5) A disk rotates about its central axis starting form rest and accelerates with constant angular
acceleration. At one instant it is rotating at 12 rad/s and after 80 radian of more angular
displacement, its angular speed becomes 28 rad/s. How much time (seconds) does the disk takes to
complete the mentioned angular displacement of 80 radians.

6) A particle is moving with constant speed in a circular path. The ratio of average velocity to its

instantaneous velocity when the particle describes an angle is to

 PART-2 : CHEMISTRY

SECTION - I (i)

1) Statement-1 : is linear while is bent.

Statement-2 : Cl - atom in and in , is having same state of hybridisation.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
(B)
statement‑1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.

2) The order of bond length from the following structure.

(A) III > I = V > IV > II

(B) IV > II > I = V > III
(C) III > IV > I = V > II
(D) I > II > III > IV > V

3) Moles of ABC3 produced in the following set of reaction when 180 gm of A, 180gm of B and 200
gm of C are mixed in a container (given molar masses of A,B,C are 20, 30 & 10 respectively.
2A + 3B + 5C → A2BC2 + B2C3
B2C3 + 3C → 2BC3
BC3 + A → ABC3

(A) 5
(B) 4
(C) 10/3
(D) 20/3

4) Consider the following reaction

2Na + 2NH3 → 2NaNH2 + H2(g)
2NaNH2 + C → Na2CN2 + 2H2 (g)
Na2CN2 + C → 2NaCN
51gm dry ammonia gas is passed over excess heated sodium to form sodamide (NaNH2) which is
further reacted with carbon (excess) to finally form NaCN. Find the total volume of H2(g) evolved at
0.5 atm, 273K

(A) 201.6 L
(B) 100.8 L
(C) 403.2 L
(D) 50.4 L

SECTION - I (ii)

1) 5.6 g N2(g) and 0.6 g H2(g) is taken to react and form NH3(g).
N2(g) + 3H2(g) → 2NH3(g)
CORRECT statements is/are -

(A) H2(g) is limiting reactant

(B) 2.8 g N2(g) is left
(C) 3.4 g NH3(g) is formed
(D) 1.7 g NH3(g) is formed

2) incorrect order of bond angle is :

(A)
(B)

(C)

(D)

3) The structure of XeF6 is :

(A) pentagonal bipyramidal

(B) distorted octahedral
(C) capped octahedral
(D) square pyramidal

SECTION - I (iii)

Common Content for Question No. 1 to 2

Hybridisation is a theoretical concept, as state of hybridisation cannot be detected even by
spectroscopically; unlike intermediates or transition state in various reactions. but it corrects the
predictions which are based simple on overlapping of pure atomic orbitals. VSEPR theory predicts
precisely shape and bond angle in a given molecule.

1) The correct order of energy levels of hybrid orbitals.

(A) sp > sp2 > sp3

(B) sp < sp2 < sp3
(C) sp2 > sp3 > sp
(D) sp3 > sp > sp2

2) In which pair of molecules bond angles are not same :-

(A) CCl4 & SiCl4

+ +
(B) NH4 & NF4
+
(C) ClF6 & SF6
(D) None of these

Common Content for Question No. 3 to 4

9 × 1022 atoms of Ar and n moles of O2 are kept in a vessel of capacity 5 L at 1 atm and 27°C.
(Consider NA = 6 × 1023) :

3) Find the mass of O2 in vessel -

(A) 16 gm
(B) 3.2 gm
(C) 1.6 gm
(D) 32 gm

4) If 3 × 1022 molecules of O2 are added in vessel. Find the new pressure in vessel at same
temperature in vessel-

(A) 1.23 atm

(B) 2.23 atm
(C) 0.23 atm
(D) None of the above

SECTION - II

1) Calculate mass (in gm) of O atoms in 6 gm CH3COOH ?

2) The weight (in gram ) of pure potash Alum (K2SO4.Al2(SO4)3.24H2O) which contains 6.4 gm oxygen
is.
(Atomic weight of K = 39, S = 32, Al = 27)

3) The Kohinoor diamond was the largest diamond ever found. How many moles of carbon atom
were peresent in it, if it is weigh 3300 carat. [Given: 1 carat = 200 mg]

4) The number of hypovalent species among the following

CF4, BF3, PCl5, H2S, SF4, BeCl2, OF2, AlCl3

5) Find the number of molecules in which atleast two d-orbitals are involved in hybridisation :-
XeF5–, SF4, XeF6 , SO3F–, CF4 , SO3 , , IF7

6) Count the total number of unshared electrons on I3– = x

Count the total number of pπ-dπ bonds in
Count the total number of lone pairs in XeF4 = z
Hence, find the value of x – z – y.

PART-3 : MATHEMATICS

SECTION - I (i)

1) Let a, b are roots of the equation x2 – 3x + p = 0 and c, d are roots of the equation x2 – 12x + q =

0, where p, q ∈ R. If a, b, c, d (taken in that order) form geometric progression then equals

(A)

(B)

(C)

(D)

2) Let ƒ(n) = n2 + n + 1, then , then (a + b) is equal to

(A) 2
(B) 3
(C) 6
(D) 9

3) If A=tan6° tan42° and B=cot66° cot78°, then -

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

4) If x+y=3-cos4θ and x-y=4 sin2θ then

(A) x4+y4=9
(B)
(C) x3+y3=2(x2+y2)
(D)

SECTION - I (ii)

1) If then which of the following holds good?

(A)
tan2x =

(B)
tan2x =

(C)

(D)

2) If & , then P2 + Q2 is -

(A) rational number

(B) irrational number
(C) algebraic number
(D) can not comment

3) Let , then

(A) S9 is divisible by 4
(B) S9 is divisible by 21
10
(C) 10.S10+3 = 3
(D) 7(S7 + 3), 10(S10 + 3), 13(S13+ 3) are in G.P.

SECTION - I (iii)

Common Content for Question No. 1 to 2

Let and for sequence {Tn}, T1 = 5, n > 2, Tn > 0, Sn = ƒ(Sn–1), (where Tn

th
denotes n term of the sequence and Sn denotes sum of n terms) :

1) If value of n for which is minimum is (a and b are co-prime), then a + b is-

(A) 3
(B) 4
(C) 7
(D) 10

2) T2018 + S25 is equal to-

(A) 20300
(B) 21300
(C) 22300
(D) 23300

Common Content for Question No. 3 to 4

If sinA + sinB = and

3) The value of sin(A + B) is -

(A)

(B)

(C)

(D)

4) The value of tan(A + B) is -

(A)

(B)

(C)

(D)

SECTION - II

1) Value of is equal to

2) If x = cos α + cos β – cos(α + β) and y = 4 sin sin cos , then (x–y) equals

3) Find the number of solution of

4)
If a, b, c are positive real numbers and the minimum value of

is m, then the value of |m| is

5) is equal to

6) Let Tn be the nth term of a series for n ≥ 1 and T1 = 1, Tn + 1 = 2Tn + 4n, n ≥ 1. If Sn = Sum of first n

terms of the series and where a, b ∈ I, then the value of is

 ANSWER KEYS

PART-1 : PHYSICS

SECTION - I (i)

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

SECTION - I (ii)

Q. 5 6 7
A. A,C C,D A,B,C

SECTION - I (iii)

Q. 8 9 10 11
A. B A A C

SECTION - II

Q. 12 13 14 15 16 17
A. 25.00 32.00 17.00 9.00 4.00 2.00

PART-2 : CHEMISTRY

SECTION - I (i)

Q. 18 19 20 21
A. C C B A

SECTION - I (ii)

Q. 22 23 24
A. A,B,C C,D B,C

SECTION - I (iii)

Q. 25 26 27 28
A. B D C A

SECTION - II

Q. 29 30 31 32 33 34
A. 3.20 9.48 55.00 3.00 3.00 3.00

PART-3 : MATHEMATICS
 SECTION - I (i)

Q. 35 36 37 38
A. B D C D

SECTION - I (ii)

Q. 39 40 41
A. B,C A,C A,B,D

SECTION - I (iii)

Q. 42 43 44 45
A. A D A D

SECTION - II

Q. 46 47 48 49 50 51
A. 2.00 1.00 7.00 0.20 1.50 1.50
 SOLUTIONS

PART-1 : PHYSICS

1)
From constant relation
a1 + a2 = 2a3

2) = 6ti – 8t5j
at t = 1 sec ; = 6i – 8j

= – 50 [6i – 8j]
= 50 [8j – 6i]
= 100 [4j – 3i]

3)

2TVb = 3TVa

Va =

4)

Speed after t time v = at t

ac = at

= 2.

=2
t = 1sec
5)
T = m1g

when thred is burnt, tension in spring remains same = m1g.

m1g – m2g = m2a g = a = upwards

for m1 a=o

6)
F.B.D. of trolley is w.r.t. ground
F.B.D. of suspended mass is w.r.t. Trolley.
Tcos 37° – mg = 0 [Equilibrium of mass in y direction w.r.t. trolley]

⇒T= T = 25 N
T sin 37° – ma = 0 [Equilibrium of mass in x direction w.r.t. trolley]

⇒a= =
F – T sin37° = 8a [Newton’s II law for trolley in x direction w.r.t. ground]

⇒F=8× + 25 × F = 75 N

7)

Ans. (A,B,C)
A,B,C is the correct answer

8)

Ans. (B)
B is the correct answer

9) (55 to 57) x = a sin ωt

y = a – a cos ωt

|v| = aω

ax = –aω2 sin ωt
ay = aω2 cos ωt

10) Let the acceleration of b is in upwards direction thay

a cos 37° = b sin 37°

11)
Balacing force for A in Hz direction
F = N' sin 37° ⇒

...... (i)
Balancing force for B in vertical direction
N'cos 37° + N3 = mg
N' cos 37° + N3 = mg
for lifting the black B N3 = 0
N' cos 37° = mg ...... (ii)

Putting value of N' from (i)

12)
N = ma .......(1)
f = mg
f ≤ fs max
mg ≤ µN
mg ≤ µma

a≥
a ≥ 25

amin =

13)
10g – 80 = 10a
a = 2 m/s2
T + 60 – 80 = 6a
T = 6 × 2 + 20
T = 32N
 14)
F = 0.4 × 2 × 10 = 8 N
F – 8 = 6 × 1.5
F = 17 N

15) FBD of 2 kg block

N + 15 – 20 = 2 × 2
⇒ N = 5 + 4 = 9 Newtons

16)

17)

PART-2 : CHEMISTRY

18)
19)
decided by
I > II size of Surrounding Atom
I < IV size of Hybrid orbital
I=V same
III > I &V size of Surrounding Atom
III > IV size of Surrounding Atom

20)

21)

Total moles of H2 = 4.5 mole

volume of H2 =

22)
 N2(g) + 3H2(g) → 2NH3(g)
Initial 0.2 mole 0.3 mole

Ratio >
(mole/SC)
H2 is L.R.
Final 0.1 mole 0 mole 0.2 mole

23) Incorrect order

(A)OCl2 >SF2 → correct
(B) H2 O > OF2 → correct

(C) > CF4 → incorrect → because equal bond angle

(D) NF3 > NH3 → incorrect, as NH3 > NF3

24)

25)

26) CCl4, SiCl4 , NH4+, NF4+ } all having θ = 10928

ClF6+ & SF6 } all bond angle are same.

27) ∴ n = 0.05 ∴ mass = 1.6 gm

28) Earlier moles = 0.2, new moles = 0.25 ∴ Pressure =

29)
60 gm CH3COOH has 32 gm O
6 gm CH3COOH will hvae 3.2 gm O-atoms

30) → has 40 atoms of O.

640 gm O is present in 948 gm alum
6.4 gm O will be present in 9.48 gm alum.

31) Weight of carbon =

Moles of carbon =

32)

Ans. (3.00)
(3.00) is the correct answer

33)

XeF5 → sp3d3
SF4 → sp3d

XeF6 → sp3d3
→ sp3
CF4 → sp3
SO3 → sp2
→
sp3d

IF7 → sp3d3

34)

y=1

z = 14
PART-3 : MATHEMATICS

35) a + b = 3 ; ab = p
c + d = 12 ; cd = q
Let a = a,
b = ar
c = ar2
and d = ar3

∴
⇒ r = ±2

so,

36)

37) We have

Multiply & divide by tan54°

⇒ tan6°tan42°tan66°tan78° ×

⇒
[∴ tan(60° – θ)tanθ tan(60° + θ) = tan3θ]

⇒
⇒A=B

38) On adding and subtracting

x=
y=
x = 2 (1 + sin2θ ) – cos22θ ;
y = 2 (1 – sin2θ) – cos22θ
x = 1 + 2 sin2θ + sin22θ ;
y = 1 – 2 sin2θ + sin22θ
x = (1 + sin2θ)2 ;
y = (1 – sin2θ)2

[Alternate : Or put θ = and verify

39)

⇒
⇒ 8(5 sin4x + 3 + 3 sin4x – 6 sin2x) = 15
⇒ 64sin4x – 48 sin2x + 9 = 0
(8 sin2x – 3)2 = 0

⇒ sin2x = and cos2x =

∴ tan2x = ⇒ (B)

and sin6x = and cos6x =

⇒ ⇒ (C)

40)

&

41)
(A) (B) S9 = 37 – 3 = 3(272 – 1) = 3.26.28
(C)
(D) n(Sn + 3) = 3n
so 37, 310,313 are in G.P.

42)

n can't be
⇒ either ans. is 2 or 3
value at 2 is –15

value at 3 is
Hence minimum value occure at n = 2

⇒
⇒a+b=3

43)
Sn = ƒ(Sn–1)

⇒ T2 = 15
⇒ T3 = 25
Tn = 10n – 5
Sn = 5n2 T2018 + S25 = 23300

44) sinA + sinB = ,

.....(1)

....(2)

⇒
45)

46) =

47)

x = cos α + cos β – cos (α + β)

y = 4 sin sin cos

x–y = cos α + cos β – cos (α + β) – 4 sin sin cos

= cos α + cos β – cos (α + β) – 2 cos

= cos α + cos β – cos (α + β) + 2 cos2 – 2 cos cos = 1 Ans.

48) Let f(x) = sinx and

f(x) = sinx and

intersect at seven points

49)

Let a + c + 2b = x .... (1)

a + b + 2c = y .... (2)
a + b + 3c = z .... (3)
c = z – y; b = x + z – 2y
a = –x + 5y – 3z.
50)

Let Sn = 1 + 4 + 10 + 20 + 35 + 56 +..... + Tn
Sn = 1 + 4 + 10 + 20 + 35 +..... + Tn
Tn = 1 + 3 + 6 + 10 + 15 + 21 +..... + tn
Tn = 1 + 3 + 6 + 10 + 15 +..... + tn
tn = 1 + 2 + 3 + 4 +..... n terms

Now

51)

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