10-08-2025

3010CJA101001250006 JA

PART-1 : PHYSICS

SECTION-I(i)

1) In the figure shown, all strings are massless. Choose the correct option(s) out of the following.

(A) T1 > T3
(B) T3 > T1
(C) T2 > T1
(D) T2 > T3

2) In the arrangement shown in figure all surfaces are smooth. Select the correct alternative(s)

(A) For any value of θ acceleration of A and B are equal

(B) Contact force between the two blocks is zero only if mA < mB
(C) Contact force between the two blocks is zero for any value of mA or mB
(D) Normal reactions exerted by the wedge on the blocks are equal

3) A 5 kg block has a uniform rope of mass 2 kg attached to its underside and a 3 kg block
is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an
external force F0. (g = 10 m/s2)

(A) The value of F0 is 120 N

(B) the net force on rope is 4 N
(C) the tension at middle point of the rope is 48 N
(D) None of these

4) Force acting on a block versus time graph is as shown in figure. Choose the correct options. (g =

10 ms–2)

(A) At t = 2 s, force of friction is 2 N

(B) At t = 8 s, force of friction is 6 N
(C) At t = 10 s, acceleration of block is 2 ms–2
(D) At t = 12 s, acceleration of block is 4 ms–2

5) Refer the system shown in the figure. Block is sliding down the wedge. All surface are frictionless.

Find the correct statement(s)

(A) Acceleration of block is g sinθ

(B) Acceleration of block is g cosθ
(C) Tension in the string is mg cos 2θ
(D) Tension in the string is my sin θ.cosθ

6) In the given diagram frame is moving with constant acceleration 4 m/s2. The string make an angle
θ with the vertical in equilibrium. If mass of body is 10kg then (g = 10 m/s2)
(A)
Value of tan

(B)
Value of
(C) Tension of string is
(D) Tension of string is 60 N

7) In the given figure both the blocks have equal mass. When the thread is cut, which of the
following statements give correct description immediately after the thread is cut?

(A) Relative to the blocks, acceleration of block B is 2g upwards

(B) Relative to the block B, acceleration of blocks is 2g downwards
(C) Relative to the ground, accelerations of the blocks A and B are both g downwards
Relative to the ground, accelerations of the blocks, 4 and B are 2g downwards and zero
(D)
respectively

8) A block of mass 1kg is held at rest against a rough vertical surface by pushing by a force
horizontally. The coefficient of friction is 0.5. When

(A) F = 40 N, friction on the block is 20 N.

(B) F = 30 N, friction on the bock is 10 N
(C) F = 20 N, friction on the block is 10 N
(D) Minimum value of force F to keep block at rest is 20 N

SECTION-I(ii)

Common Content for Question No. 1 to 2

Answer the following by appropriately matching the lists based on the information given in
the paragraph.

In shown situation elevator is moving upward with acceleration of 5 m/s2.

List-I List-II

(I) Net force acting on B (P) 350 N

(II) Normal reaction between A and B (Q) 300 N

(III) Normal reaction between B and C (R) 200 N

(IV) Normal reaction between C and elevator (S) 750 N

(T) 1500 N

(U) 150 N

1) If mA = 10 kg, mB = 40 kg, mC = 50 kg then choose the correct match

(A) (I) → (R)

(B) (II) → (T)
(C) (III) → (U)
(D) (IV) → (P)
2) If mA = 20 kg, mB = 30 kg, mC = 50 kg then choose the correct match

(A) (I) → (T)

(B) (II) → (P)
(C) (III) → (S)
(D) (IV) → (U)

Common Content for Question No. 3 to 4

Match the list – I with list – II.

3) Which of the following is only correct combination.

(A) III → Q; IV → T
(B) III → R; IV → Q
(C) III → Q; IV → R
(D) III → T; IV → R
4) Which of the following is only correct combination.

(A) I → P; II → S
(B) I → S; II → P
(C) I → Q; II → P
(D) I → S; II → Q

SECTION-II

1)

A block of mass 20 kg is held stationary by a man of mass 50 kg as shown in figure. The action on
the floor by the man is α × 102 N. Fill the value of α.

2) If the tension in the string in figure is 4 N and all the surfaces are frictionless. Find the value of

M.

3) In the given figure a block of mass 8 kg is placed on a rough horizontal surface and is acted upon
by two forces F1 and F2. The magnitude of acceleration of the block is.

4) In the given figure threads and pulleys are ideal. Masses of the blocks are m1 = 1kg and m2 =1 kg
then acceleration of m1 in m/s2 will be (g =10 m/s2)

5) Two weights W1 and W2 in equilibrium and at rest are suspended as shown in figure. Then find

ratio of . Strings and pulley are ideal.

6) Figure shows a man of mass 48 kg standing on a light weighing machine kept in a box of mass 30
kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which
is held by the man himself. If the man manages to keep the box at rest, the weight shown by the

machine is ______kg.

PART-2 : CHEMISTRY

SECTION-I(i)

1) Lone pair of central atom occupies the equitorial position in

(A) ClF3
(B) XeF2
(C) SF4
(D) None of these

2) Which of the following orders of atomic / Ionic radius is correct?

(A) B < Ga ≲ Aℓ
(B) Cu < Zn
(C) C < O < N
(D) Al3+ < Al2+ < Al+

3) Choose the correct statement(s)

(A) In Moseley's equation, square root of frequency is directly proportional to atomic number
(B) Silver and gold have nearly same atomic radius
(C) Sc <Y< La (correct order of atomic size)
(D) Transition elements show variable oxidation states.

4) 24.6 eV is required to remove one of the electrons from a helium atom. The energy required to
remove both electrons from helium atom is:

(A) 38.2 eV/atom

(B) 49.2 eV/atom
(C) 54.4 eV/atom
(D) 79 eV/atom

5) Moseley equation is represented as , where a and b are constant. If OA = 1,

calculate nuclear charge Z. Where frequency is 400 Hz.

(A) 21
(B) 20
(C) 40
(D) 41

6) Given

2ZnS + 3O2 2ZnO + 2SO2

ZnO + H2SO4 ZnSO4 + H2O

2ZnSO4 + 2H2O 2Zn + 2H2SO4+O2

If 15 mole of ZnS reacts with excess of reactants then which of the following statement(s) is/are
correct?
(A) 15 mole of Zn formed
(B) 12 mole of SO2 gas released after first reaction
(C) 4.5 mole of H2SO4 formed
(D) 9 mole of ZnSO4 formed after second reaction

7) A definite volume of pure ammonia (NH3) gas is passed through a series of electric sparks by
which the volume becomes 90 ml. The increase in volume is due to formation of nitrogen (N2) and
hydrogen (H2) gases. All the gases finally present are washed with dilute sulphuric acid solution, by
which the volume of gases becomes 80 ml. All the volumes are measured at the same temperatures
and pressure. Which of the following statement(s) is/are correct regarding the original ammonia
sample?

(A) The volume of NH3 gas taken was 40 ml.

(B) The volume of NH3 gas taken was 50 ml.
(C) Only 80% of NH3 gas decomposed into N2 and H2 gases.
(D) Only 20% of NH3 gas decomposed into N2 and H2 gases.

8) Which of the following statement are correct for P4S3 :

(A) No. of P – S – P Bonds are 3

(B) No. of P – P Bonds are 3
(C) No. of P = S Bonds are 3
(D) Number of lone pairs present = 10

SECTION-I(ii)

Common Content for Question No. 1 to 2

List-I List-II

(P) (1) All bond lengths are same

(Q) (2) All bond angles are same

Maximum number of atoms

(R) (3)
in one plane are ≥4

(S) (4) Non-planar

(5) 2pπ–3dπ bonding is present

(6) Octahedral shape

1) Which of the following is correctly matched?

(A) P–1–4
(B) S–3–2
(C) R–2–6
(D) P–3–4
2) Which of the following is INCORRECTLY matched?

(A) P–3–4
(B) S–1–4
(C) Q–4–6
(D) R–1–4

Common Content for Question No. 3 to 4

Match the List.

List-I List-II
(P) H4P2O7 (1) Basicity ≥ 3
(Q) H2S2O8 (2) Peroxy linkage
(R) H3P3O9 (3) Cyclic structure
X – O – X linkage where
(S) H3B3O6 (4)
(X = P, S, B)
(5) Meta acid

3) Choose the correct matched option

(A) S–1–4–3
(B) P–2–1–3
(C) R–2–1–4
(D) Q–1–2–4

4) Which is INCORRECTLY matched?

(A) P–1–4
(B) Q–2–4
(C) R–1–4–3
(D) S–3–4

SECTION-II

1) The sum of the five and six member rings in C60 is.

2) Total number of compound having 3-D network like structure. Diamond, CO2, SiC, Graphite,
Borazine

3) Central atom may exhibit sp3 hybridisation in how many of the following species :
(a) CO2 (b) Graphite
(c) Diamond (d) CO
(e) H3BO3 (f) H3P3O9
(g) S3O9 (h) (SiO4)4–
(i) H2B4O7 (j) PCl5(s)
(k) I2Cl6 (l) (l) Perchloric Acid
(m) H2CO3 (n) COCl2
(o) Sulphite ion (p) CCl4

4) Nitric acid can be produced in three steps process :

(i) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(ii) 2NO(g) + O2(g) → 2NO2(g)
(iii) 3NO2(g) + H2O(ℓ) → 2HNO3(aq.) + NO(g)
% yield of Ist, IInd and IIIrd reaction are respectively 50%, 60% and 80%, and the volume of NH3(g) at
1 atm and 0°C required to produce 1575g of HNO3 is v (in litre). Then find v/100 ?

5) In how many following pairs, first species has higher ionisation energy than second species :
(i) Na+, Mg2+ (ii) S, Cl (iii) Cu, Zn (iv) Xe, Kr
2–
(v) B, Be (vi) O , O (vii) Al, Si (viii) Cl–, Cl

6) In 2 moles of KHC2O4.Na2.C2O4.3H2O there are 'X' mole atoms of oxygen and 'Y' mole atoms of
Hydrogen then what is X – Y = ..

PART-3 : MATHEMATICS

SECTION-I(i)

1) If b2 ≥ 4ac for the equation ax4 + bx2 + c = 0 then all the roots of the equation will be real if

(A) b > 0, a < 0, c > 0

(B) b < 0, a > 0, c > 0
(C) b > 0, a > 0, c > 0
(D) b > 0, a < 0, c < 0

2) The adjoining figure shows the graph of y = ax2 + bx + c. Then

(A) a > 0
(B) b > 0
(C) c > 0
(D) b2 < 4ac
3) Consider the equation :

and S be the solution set of above

equation, then

(A) Sum of absolute values of real solution(s) of given equation is a prime number
(B) equation has more than two real solutions

(C)

(D) Sum of the roots is –3

4) The equation acosx – cos2x = 2a – 7 can have a solution, if

(A) a = 2
(B) a = 6
(C) a ∈ (2, 6)
(D) a ∈ [2, 6]

5) If then tan can have the value equal to-

(A) 2
(B) 1/2
(C) –2
(D) –1/2

6) Solution set of the of the equation sin 2θ = cos 3θ is/are.

(A)

(B)

(C)

(D)
(where n ∈ I)

7) 7 points are given in a plane (as shown in adjacent figure), then using these 7 points –

(A) Number of lines that can be formed is 10.

(B) Number of triangles that can be formed is 24.
(C) Number of quadrilateral that can be formed is 6.
Maximum number of circles that can be formed is 24, where each circle contains atleast three
(D)
of 7 points.

8) In how many ways 5 boys and 7 girls are seated in a row so that boys are separated ?

(A)

(B)

(C)

(D)

SECTION-I(ii)

Common Content for Question No. 1 to 2

Let f(θ) = sinθ + cosθ and g(θ) = sin2θ + cos2θ where θ ∈ [0, 2π]. Define the following sets with
elements listed in increasing order.
X = {θ ∈ [0, 2π] : f(θ) = }
Y = {θ ∈ [0, 2π] : f(θ) = 0}
Z = {θ ∈ [0, 2π] : g(θ) = 0}
W = {θ ∈ [0, 2π] : g(θ) = 1}

List-I List-II

(I) X (P)

(II) Y (Q)

(III) Z (R)

(IV) W (S) Arithmetic Progression

(T)

(U)

1) Which of the following is the only correct combination

(A) (I)-(P) ; (II)-(T) ; (III)-(Q)

(B) (II)-(Q) ; (IV)-(U) ; (III)-(R)
(C) (I)-(P) ; (IV)-(U) ; (III)-(Q)
(D) (I)-(T) ; (II)-(Q) ; (IV)-(S)

2)
Which of the following is the only correct combination ?

(A) (III)-(QS)
(B) (IV)-(PRU)
(C) (III)-(PQU)
(D) (IV)-(QT)

Common Content for Question No. 3 to 4

Let f(x) = (a – 5)x2 – 2ax + (a – 4), where a ∈ R, a ≠ 5.
Define the following sets of values of a :
* P. All values of a for which the product of roots of f(x) is positive
* Q. All values of a for which the product of roots f(x) is negative
* S. All values of a for which the roots of f(x) are real
* T. All values of a for which f(x) has complex roots
Let List-I contain the sets P, Q, R, S, T and List-II contain correct characterizations of these sets.

List-I List-II

(i) P (A)
– {5}

(ii) Q (B) 5

(iii) S (C)

(iv) T (D)

(E)

(F) (4,5)

3) Which of the following is the only correct combination

(A) (i)-(A) ; (ii)-(B) ; (iii)-(C) ; (iv)-(D)

(B) (i)-(D) ; (ii)-(B) ; (iii)-(C) ; (iv)-(A)
(C) (ii)-(D) ; (ii)-(F) ; (iii)-(A) ; (iv)-(C)
(D) (ii)-(D) ; (ii)-(F) ; (iii)-(C) ; (iv)-(A)

4) Which of the following is the only correct combination

(A) (ii)-(AB)
(B) (iii)-(CD)
(C) (iv)-(AB)
(D) (ii)-(EF)

SECTION-II

1) Your math club has 20 members. In how many ways can it select a president, a vice-president,
and a treasurer if no member can hold more than one office ?

2) If the number of arrangement of 4 alike apples, 5 alike mangoes, 1 banana and 1 orange in which
all the apples are together or all the mangoes are together is K, then find the sum of digits in K.

3) If for some p, q, r ∈ R, not all have same sign, one of the roots of the equation

(p2 + q2) x2 –2q (p + r) x + q2 + r2 = 0 is also a root of the equation x2 + 2x – 8 = 0, then is

equal to-

4) The value of k for which 2x2 + 7xy + 3y2 + 8x + 14y + k = 0 can be resolved in to two linear
factors is :

5) If cot = , (where a, b, c, d ∈ Ι+) then value of a + b + c + d is

6) Find the number of values of θ satisfying the equation sin3θ = 4sin θ. sin 2θ. sin 4θ in 0 ≤ θ ≤ 2π
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I(i)

Q. 1 2 3 4 5 6 7 8
A. B,C,D A,C A,B,C A,B,C A,D A,C A,B B,C,D

SECTION-I(ii)

Q. 9 10 11 12
A. A C C B

SECTION-II

Q. 13 14 15 16 17 18
A. 3.00 4.00 5.00 4.00 1.25 9.00

PART-2 : CHEMISTRY

SECTION-I(i)

Q. 19 20 21 22 23 24 25 26
A. A,B,C A,B,D A,B,C,D D A B,C,D B,C A,B,D

SECTION-I(ii)

Q. 27 28 29 30
A. D D A B

SECTION-II

Q. 31 32 33 34 35 36
A. 32.00 3.00 10.00 35.00 0.00 8

PART-3 : MATHEMATICS

SECTION-I(i)

Q. 37 38 39 40 41 42 43 44
A. B,D B,C A,C,D A,B,C,D A,B,C,D A,B,C A,B,C,D A,C

SECTION-I(ii)

Q. 45 46 47 48
A. A A C D
 SECTION-II

Q. 49 50 51 52 53 54
A. 6840.00 9.00 272.00 8.00 15.00 15.00
 SOLUTIONS

PART-1 : PHYSICS

1)

Answer : (BCD)

2)

Answer : (AC)

3)

Answer : (ABC)

4)

Answer : (ABC)

5) Answer : (AD)

6)

Answer : (AB)

7)

(A) Relative to the blocks, acceleration of block B is 2g upwards

(B) Relative to the block B, acceleration of blocks is 2g downwards

8)

Answer : (BCD)

9)

For (I) : Net force on B = ma = (40) (5) = 200N

For (II) : FBD of A NAB – 10 g = 10(5) ⇒NAB = 150N

For (III) : FBD of B NBC –NAB – 40 g = 40(5) ⇒NBC = 750N
For (IV) : For system of A + B + C : NC – 100 g = 100 (5)⇒ NC = 1500 N

10)

(III) → (S)

11) Ans. I → S; II → P; III → Q; IV → R

12) Ans. I → S;II → P;III → Q;IV → R

13) Consider the forces on the man in equilibrium : his weight, force due to the rope and
normal force due to the floor.

14) Ans. 4

Sol.

a = 4 m/s2
mg sin 30° – 4 = ma
5m – 4 = 4m
m = 4kg

15)
N = 40N
fk = µN = 8

16)

Answer : 4

17) W1 cos37° = W2
 18)
2T = (m1 + m2)g
2T = 78g
T = 39g

N + T = 48g
N = (48 – 39)g
N=9g
Hence 9 kg wt.

PART-2 : CHEMISTRY

19)

Answer (ABC)

20)

(A) B < Al ≅ Ga ⇒ B < Al : n ,size

⇒ Al ≅ Ga : d-contraction/scandide contraction
(B) Cu < Zn ⇒ 3d- series size order
(C) C < O < N ⇒ incorrect order as left to right in period, size decreases
Correct order: C > N > O
(D) Al3⊕ < Al2⊕ < Al⊕ ⇒ As cationic charge increases on atom, zeff increases & hence size of
cation decreases.

21)

Answer (ABCD)

22)

Answer (D)

23)

24)

2ZnS + 3O2 2ZnO + 2SO2

15 mole 15 × 0.8 = 12 mole ZnO

ZnO + H2SO4 ZnSO4 + H2O

12 mole 12 × 0.75 = 9 mole ZnSO4

2ZnSO4 + 2H2O 2Zn + 2H2SO4 + O2

9 mole 9×0.5 9×0.5 4.5×0.5
= 4.5 mole = 4.5 mole = 2.25 mole

25)

Answer : (BC)

26) Answer (ABD)

No. of P – S – P Bonds are 3
No. of P – P Bonds are 3
Number of lone pairs present = 10

27) Answer (D)

P–3–4

28) Answer (D)

R–1–4

29) Answer (A)

 S–1–4–3

30) Answer (B)

Q–2–4

31)

Answer (32)

32) Answer (3)

Diamond, SiO2, Graphite

33) Answer (10)

(a)

34)

Moles of NO2 required

Moles of NO required =

Moles of NH3 required =

= 156.25
Volume of NH3 at 1 atm and 0°C required is
= 156.25 × 22.4
= 3500 L

35)

Answer (0)

36) Answer (8)

In 2 moles of KHC2O4.Na2.C2O4.3H2O there are 'X' mole atoms of oxygen and 'Y' mole atoms of
Hydrogen then what is X – Y = 8

PART-3 : MATHEMATICS

37) Answer (BD)

b < 0, a > 0, c > 0
b > 0, a < 0, c < 0

38) By seeing graph of ax2 + bx + c = y

c > 0 and b > 0
So ans is (B) and (C).

39) Answer (ACD)

Given equation is x2 = 4 − 3x x = 1, −4

40) acosx – cos2x = 2a – 7

acosx – (2cos2 x – 1) = 2a – 7
acosx – 2cos2 x = 2a – 8
2cos2 x – acosx + 2a – 8 = 0

41)

Answer : (ABCD)

42)

43) (A) 7C2 – 3C2 – 5C2 + 2 = 10

(B) 7C3 – 3C3 – 5C3 = 24
6
(C) there is no use of point '3' C4 – 4C4 – 4C3.2C1 = 6
(D) 7C3 – 3C3 – 5C3 = 24

44)

45) (I)-(P) ; (II)-(T) ; (III)-(Q)

46)

Answer : (A) (III)-(QS)

47) (ii)-(D) ; (ii)-(F) ; (iii)-(A) ; (iv)-(C)

48) Answer : (ii)-(EF)

49)

Answer : 6840

50)

Answer : 9

2 2
51) (px – q) + (qx – r) = 0

52) Δ = abc + 2fgh – af2 – bg2 – ch2 = 0

⇒k=8

53)

Answer : 15

54) sin 3θ = 4 sin θ sin 2θ sin 4θ ⇒ sin 3θ = (2 sin θ) (2 sin 2θ sin 4θ)
⇒ 3 sin θ – 4 sin3 θ = 2 sin θ (cos 2θ – cos 6θ) ⇒ 3 – 4 sin2 θ = 2(cos 2θ – cos 6θ) or sin θ = 0
⇒ 3 – 2(1 – cos 2θ) = 2 cos 2θ – 2 cos 6θ or sin θ = 0

⇒ 1 = –2 cos 6θ ⇒ cos 6θ = or sin θ = 0 ∴ sin θ = 0 or cos 6θ =

⇒ θ = nπ or θ = = ⇒ θ = 0, π, , , ,
So eight solutions.