18-05-2025

2001CJA101031250001 JA

PHYSICS

SECTION-I

1) In the given figure

(A) Angle between and is 110°

(B) Angle between and is 60°
(C) Angle between and is 110°
(D) Angle between and is 70°

2) A force of 4N and another of 9N can be applied together to produce the effect of a single force of :

(A) 1 N
(B) 11 N
(C) 15 N
(D) 20 N

3) The vector sum of two vectors and is maximum, then the angle θ between two vectors is :

(A) 0º
(B) 30º
(C) 45º
(D) 60º

4) If and are two non–zero vectors such that = and = then the angle
between and is :

(A) 37º
(B) 53º
(C) cos–1(–3/4)
(D) cos–1(–4/3)

5) In an octagon ABCDEFGH of equal side, what is the sum of

if,
(A)
(B)
(C)
(D)

6) A train is accelerating uniformly passes by a stationary pole. If the time of crossing the first half
length of the train is twice that the time in crossing the remaining half. The speeds of the front of
the engine, middle point and the back of train when it crosses the pole are in the ratio.

(A) 1 : 2 : 3
(B) 1 : 3 : 5
(C) 1 : 3 : 7
(D) 1 : 5 : 7

7) A projectile of mass m is fired with velocity v from the point P at an angle 45° with the horizontal.
The magnitude of change in momentum when it passes through the point Q on the same horizontal
line on which P lies is :

(A)

(B)

(C) 0
(D) 2mv

8) A particle is projected at an angle θ above the horizontal with a speed u. After some time the
direction of its velocity makes an angle ϕ above the horizontal. The speed of the particle at this
instant is

(A)

(B)

(C)

(D)
9) If the velocity -time graph has the shape AMB, what would be the shape of the corresponding

acceleration - time graph?

(A)

(B)

(C)

(D)

10) A particle is projected from a horizontal plane (x-z plane) such that its velocity vector at time t is
given by . Its range on the horizontal plane is given by

(A)

(B)

(C)

(D) None

11) A particle's position as a function of time is described as y(t) = 2t2 + 3t + 4. What is the average
velocity of the particle from t = 0 to t = 3 sec?

(A) 3 m/sec.
(B) 6 m/sec.
(C) 9 m/sec.
(D) 12 m/sec.

12) A man is crossing a river flowing with velocity of 5 m/s. He reaches a point directly across at a

distance of 60 m in 5 sec. His velocity in still water should be :-

(A) 12 m/s
(B) 13 m/s
(C) 5 m/s
(D) 10 m/s

13) A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first
10 s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as :-

(A) 1 : 2 : 4
(B) 1 : 2 : 5
(C) 1 : 3 : 5
(D) 1 : 3 : 9

14) The velocity at the maximum height of a projectile is half of its initial velocity u. Its range on the
horizontal plane is:

(A)

(B)

(C)

(D)

15) A particle moves with constant acceleration in the positive x-axis. At t=0, the particle is at origin
and is at rest, then correct graph between velocity and displacement is:

(A)

(B)

(C)
(D)

16) Assertion : A body can have acceleration even if its velocity is zero at a given instant of time.
Reason : A body is momentarily at rest when it reverses its direction of motion.

(A) Both assertion and reason are true and the reason is the correct explanation of the assertion.
(B) Both assertion and reason are true but reason is not the correct explanation of the assertion.
(C) Assertion is true but reason is false.
(D) Assertion and reason both are false.

17)

The block of mass m1 remains stationary if :-

(A)

(B)

(C)

(D)

18) When a constant force is applied to a body, it moves with uniform :

(A) acceleration
(B) velocity
(C) speed
(D) momentum

19) A string of length L and mass M is lying on a horizontal table. A force F is applied at one of its
ends. Tension in the string at a distance x from the end at which force is applied is:

(A) zero
(B) F
(C) F (L – x) /L
(D) F (L– x)/M

20) Two blocks are pulled upward by a constant force F as shown in the figure What is
tension in the string connecting two blocks ?

(A)

(B)

(C)

(D)

SECTION-II

1) A displacement vector, with an angle of 30° with y-axis has x-component of 10 units, then the
magnitude of the vector is 4 × n. Find the value of n.

2) and then is -

3) A projectile fired with initial velocity u at some angle θ has a range R. If the initial velocity be
doubled at the same angle of projection, then the range will be times

4) The displacement time [x – t] graphs of two particles A and B are straight lines making angles of
respectively 60° and 30° with the time axis. If the velocity of A is vA and that of B is vB, then find the

value of vA/vB

5) A block of mass 2 kg is given a push for a moment horizontally and then the block starts sliding
over a horizontal plane. The graph shows the velocity-time graph of the motion. The magnitude of
force acting on particle in newton will be.

CHEMISTRY

SECTION-I

1) A sample of ammonium phosphate (NH4)3 PO4 contains 6 moles of H-atoms. The number of moles
of O-atoms present in the sample is.

(A) 4
(B) 2
(C) 5
(D) 3

2) A gaseous mixture contains O2 & SO2 in ratio of 1 : 4 by mass. Therefore, the ratio of their
respective number of molecules is :

(A) 1 : 2
(B) 1 : 4
(C) 2 : 1
(D) 4 : 1

3) If the mass of 1022 molecules of a hydrocarbon is about 1.2g. Then Gram molecular mass of
hydrocarbon is : (Take NA = 6 × 1023)

(A) 36 g
(B) 72 g
(C) 54 g
(D) 90 g

4) A compound of nitrogen & oxygen was found to contain 7 : 16 by mass N & O respectively.
Calculate molecular formula of compound if Vapour density of compound is 46.

(A) NO2
(B) N2O4
(C) N2O3
(D) N2O5
5) A sample of copper sulphate pentahydrate, CuSO4.5H2O contains 3.782 g of Cu. How many grams
of oxygen are in this sample (Atomic mass Cu = 63.5, S = 32)?

(A) 0.952 g
(B) 3.809 g
(C) 4.761 g
(D) 8.576 g

6) Which sample contains greatest number of atoms:

(A) 2 g C (atomic weight = 12)

(B) 2g N (atomic weight = 14)
(C) 2g O (atomic weight =16)
(D) 1g He (atomic weight = 4)

7) Which sample has greatest mass :

(A) 4 mole of N2
23
(B) 12.044 × 10 molecules of O2
(C) 6.022 × 1023 atoms of H
(D) 12 gm C atoms

8) Haemoglobin contains 0.25% iron by mass. The molecular mass of haemoglobin is 89600.
Calculate the number of iron atoms per molecule of haemoglobin
[Atomic mass of iron = 56]

(A) 3
(B) 4
(C) 6
(D) 5

9) The percentage composition of carbon by mass in methane is :

(A) 80%
(B) 25%
(C) 75%
(D) 20%

10) 11.2 L of gas at 1 atm pressure and 273K. temperature weighs 14.0 g. The gas could be :

(A) N2O
(B) NO2
(C) N2
(D) CO2
11) Total number of functional groups present in above
compound

(A) 7
(B) 8
(C) 9
(D) 6

12) The principle functional group in following compound for IUPAC naming.

(A) Aldehyde
(B) Ketone
(C) Amine
(D) Amide

13) Select the correct IUPAC name of

(A) Diamino ketone

(B) Diaminomethanone
(C) Aminomethanamide
(D) Methandiamide

14) Correct IUPAC name of compound is :

(A) 1,2-dimethyl cyclobutene

(B) 2,3-dimethyl cyclobutene
(C) 1,2-dimethyl cyclobut-2-ene
(D) 1,4-dimethyl cyclobutene

15)

How many carbons are present in parent carbon chain of following compound :
(A) 5
(B) 10
(C) 15
(D) 20

16)

Choose the correct IUPAC name of following compound ?

(A) 2,4-diethyl-6-methyl heptane

(B) 3,7-dimethyl-5-ethyl octane
(C) 5-ethyl-3,7-dimethyl octane
(D) 4-ethyl-2,6-dimethyl octane

17) Choose the correct IUPAC name of following compound :

(A) Bis-3,5-isopropyl heptane

(B) 2,6-Dimethyl-3,5-diethyl heptane
(C) 3-Ethyl-5-isopropyl-2-methyl heptane
(D) 3,5-Diethyl-2,6-dimethyl heptane

18) Which of the following compound has only one type (degree) of carbon ?

(A)

(B)
(C)

(D)

19) Which type of hybridisation is present in naphthaquinone is :-

(A) sp3
(B) sp2
(C) sp
(D) All of these

20) Which one of the following is heterocyclic compound ?

(A)

(B)

(C)
(D)

SECTION-II

1) Number of gram–atom in 17 g NH3 is/are :

2) Ice has a density of 0.9 gm/ml. Calculate moles of protons in 2 ml of ice cube.

3) If 5 g H2 is mixed with 14 g of nitrogen for the following reaction :

N2 + 3H2 → 2NH3
At the end, mass of H2 (in grams) left unreacted is.

4) Number of gram–atom in 17 g NH3 is/are :

5) How many type of functional groups present in following compound ?

MATHEMATICS

SECTION-I

1) If , then

(A) x = 6
(B) x = –6
(C) x = 2
(D) x ∈ ϕ

2) If A = {1,2,3,4,5,8},
B = {3,4,5,10,11,12},
C = {4,5,8,14,15},
Then (A ∩ B) ∪ (A ∩ C) is equal to
(A) {3,4,5}
(B) {3,5,8}
(C) {3,4,8}
(D) {3,4,5,8}

3) The complete solution of , is

(A) [–2, ∞)
(B) [0, ∞)
(C) (–∞, 0) ∪ (2, ∞)
(D) (–∞, 0) ∪ [2, ∞)

4) is equal to -

(A) 2
(B) 0
(C) 4
(D) 1

5) If x = logb a, y = logc b, z = loga c, then xyz (whenever exists) is

(A) 0
(B) 1
(C) 3
(D) None of these

6)

Given that 'x' is a real number satisfying then

(A)

(B)

(C) x > 7

(D)

7) The value of is : -

(A) tan 10º

(B) tan125°
(C) cot 55º
(D) cot 10º
8) If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to

(A) 1
(B) 4
(C) 2
(D) None of these

9) If and ; then :-

(A)
(B)
(C)
(D)

10) Let for k = 1, 2, 3, .... Then for all x ∈ R, the value of f4(x) – f6(x) is
equal to :-

(A)

(B)

(C)

(D)

11) The tangents of two acute angles are 3 and 2. The sine of twice their difference is -

(A)

(B)

(C)

(D)

12) Solution set of is given by

(A) (–∞, –4) ∪ [–2, –1]

(B) (–∞, –4) ∪ (–2, –1)
(C) (–4, –3) ∪ (–3, –2] ∪ [–1, )
(D) (–4, –3) ∪ (–3, –2) ∪ (–1, )

13) If x – y = a and x2 + y2 = b, then the value of x3 – y3 is :

(A)

(B)

(C)

(D) a2 + b

14) If x2 + y2 + z2 = 11 and x + y + z = 11, then value of xy + yz + zx is -

(A) 120
(B) 110
(C) 55
(D) 22

15) If then value of is

(A) 3541
(B) 2481
(C) 3121
(D) 2207

16) The value of is

(A) 9
(B) 10
(C) 11
(D) 12

17)

Number of natural number 'n' for which is a natural number, is

(A) 2
(B) 3
(C) 4
(D) 6

18) The value of cos 255° + sin 195° is :

(A)
(B)

(C)

(D)

19) The value of is

(A) 0
(B) 1

(C)

(D)

20) The value of is equal to -

(A) 1
(B) 3
(C) 9
(D) 27

SECTION-II

1) Value of is

2) If , then sum of digits of values of x satisfying the equation

3) The value of – is

4) Number of values of x satisfying is/are :

5) The smallest positive integer which satisfies > 0, is ___.

 ANSWER KEYS

PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. C B A C B D A B B B C B C B A A D A C D

SECTION-II

Q. 21 22 23 24 25
A. 5.00 6.00 4.00 3.00 4.00

CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. B A B B D D A B C C A D C D A D D B B C

SECTION-II

Q. 46 47 48 49 50
A. 4.00 1.00 2.00 4.00 4.00

MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. D D D D B D B C B D D C B C D A D A C A

SECTION-II

Q. 71 72 73 74 75
A. 0.00 9.00 1.00 1.00 4.00
 SOLUTIONS

PHYSICS

3)

For any two vectors and

for maximum value of R, θ should be 0 so that

4)

(A2 + B2 + 2AB cos θ) = (A2 + B2 – 2AB cos θ )

⇒ 3A2 + 3B2 + 10 AB cos θ = 0
or 12B2 + 3B2 + 10(2B) (B) cos θ = 0
15B2 + 20B2 cos θ= 0

cos θ =–

5)

We know,
∵ = + + + + =
By triangle law of vector addition, we can write

Now

=
=
=
=

7)

8)
u cos θ = v cos ϕ

11) Average velocity

⇒ y(0) = 4
y(3) = 31

12)
Drift zero
Vbr sin θ = Vr

Vbr cos θ = 12 m/s

Vbr sin θ = 5 m/s
Vbr = 13 m/s

16) Note : Remove 'IF' in all options by Adil Sir

17)

....(i)
2T = m1 g ....(ii)

19)

20) F – 3mg = 3ma

T – 2mg = 2ma
 21) A cos 60° = 10 ⇒ A = 20 unit

22) = 2+2+2 = 6

23)

24) = slope of X-T graphs

= tan θ

∴ vA = tan 60°

and vB = tan 30° =

CHEMISTRY

26) 12 moles of H-atoms means 1 mol of (NH4)3 PO4

So 6 moles of H-atoms =
⇒ 0.5 moles of (NH4)3PO4
So no. of moles of O-atoms = 0.5 × 4 = 2

27) Ratio of O2 : SO2 by mass is 1 : 4

∴ Let us assume the mixture contains 1g of O2
∴ It will contain 4g of SO2

∴ Ratio of number of molecules of O2 : SO2 =

 28) Mass of 1022 molecule = 1.2g

mass of 1 molecule =

mass of NA molecule =

29) N2O4

Ratio No. of atom in

No. of
by empirical
moles
mass formula

N 7 7/14 = 0.5 1

O 16 16/16=1 2

EF of compound is NO2 :
empirical forumla mass = 46
Vapour density of compound is 46
∴ Molar mass = 2 × vd = 92

∴ M.F of compound (NO2)2 = N2O4

33)

JA13PCMC01SCM

34)

CH4 = 16 g/mol

= 75%

36)

37)

(D) Amide
 38)

39)
Lowest locant is (1, 4) instead of (2, 3)

40)

41)

46) Theory based.

47)

∴ nProton =

48)

N2 + 3H2 → 2NH3

–
– 1 1

49) Theory based.

MATHEMATICS

53)
x∈(–∞, 0) ∪ [2, ∞)

56)

We have

59) x2 – 3x + 2 > 0
(x – 1) (x – 2) > 0
x ∈ (–∞, 1) ∪ (2, ∞) ....(i)
x2 – 3x – 4 ≤ 0
⇒ (x – 4) (x + 1) ≤ 0
x ∈ [–1, 4] ....(ii)
equation (i) and (ii)
x ∈ [–1,1) ∪ (2,4]

61) Let tanθ = 3 tan θ = 2

62)
(–4, –3) ∪ (–3, –2] ∪ [–1, ∞)

63) x – y = a
∴ x2 + y2 – 2xy = a2
b – a2 = 2xy

...(1)
3 3 3
Now x – y = (x – y) + 3xy(x – y)
 =

64) Use (x + y + z)2

= x2 + y2 + z2 + z(xy + yz + zx)

65)

66) Rationalise Denomenators

67)

must be an integer
⇒ n = 1,2,3,4,6,12

68) cos 255º + sin 195º = cos(180º + 75º) + sin(180º + 15º)

= –cos 75º – sin 15º =

70)

∴
∴ y=1

71)
73)

Since = –1

75)

x ∈ (–∞, –1) ∪ (1, 2) ∪ (3, ∞)

smallest positive integer = 4