04-06-2025

6001CJM20211125001 JM

PHYSICS

SECTION-I

1) The force F is given in terms of time t and displacement x by the equation

F = a cos αx + b sin βt
where a and b are the amplitudes (amplitude is the maximum displacement of a particle). The
dimensions of β/α are :

0 0 0
(A) [M L T ]
0 0
(B) [M L T–1]
0 0
(C) [M L–1T ]
0
(D) [M L1T–1]

2) SI unit of x is , SI unit of y is and then unit of z is :

(A)

(B)

(C)

(D)

3) The velocity of a freely falling body changes as gphq where g is the acceleration due to gravity
and h is the height. The values of p and q are :

(A)
,

(B)
1,

(C)
,1
(D) 1, 1

4) The velocity u (in cms–1) of a particle is given in terms of time t (in sec.) by the relation u = at +

; the dimensions of a, b and c are :

(A) a = L2, b = T, c = LT2

(B) a = TL2, b = LT, c = L
(C) a= LT–2, b = L, c = T
(D) a = L, b = LT, c = T2

5) If the energy (E),velocity (u) and force (F) be taken as fundamental quantities, then the
dimensions of mass will be :

(A) Fu–2
(B) Fu–1
(C) Eu–2
(D) Eu2

6) What are the dimensions of a and b in the relation F = , where F is force, x is distance
and t is time:

(A) ML1/2 T–2, MLT–4

(B) ML1/2T–1, MLT–4
(C) ML1/2T–2, ML2T–4
(D) Both are dimensionless

7) If the force on a particle is given as , when x is the distance, then the dimensions
of will be (A & B are constants) :

(A) M1L1/2T–2
(B) M1L–1/2T–2
(C) M1L1T–2
(D) M1L–1T–2

8) If area (A) velocity (v) and density are base units, then the dimensional formula of force can be
represented as

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

9) If x = at + bt2, where x is in metres and t is in hours (h) then the unit of b will be

(A)

(B)

(C) m

(D)
10) Find the dimensions of velocity

0 0 0
(A) [M L T ]
0 0
(B) [M L T1]
0 0
(C) [M L T–1]
0
(D) [M L1T–1]

11) If the dimension of f is (S = displacement, t = time)

0
(A) [M L–1 T3]
(B) [M1 L1 T–3]
0
(C) [M L1T–3]
0
(D) [M L–1 T–3]

12) The value of universal gravitational constant is The value of G in

units of is

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

13) The equation of state of some gases can be expressed as . Here, P is the
pressure, V the volume, T the absolute temperature, and a, b and R are constants. The dimensions of
a are

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

14) Which among the following is a fundamental physical quantity.

(A) Speed
(B) Velocity
(C) Force
(D) Electric current

15) If velocity [V], time [T] and force [F] are chosen as base quantities, the dimensions of mass will
be

(A)
(B)

(C)

(D)

16) The density of a material A is 1500 kg/m3 and that of another material B is 2000 kg/m3. It is
found that the heat capacity of 8 volumes of A is equal to heat capacity of 12 volumes of B. The ratio
of specific heats of A and B will be

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

17) Two rods of the same length, have radii in the ratio 3:4. Their densities are respectively 8000
and 9000 kg/m3. Their specific heats are in the ratio of 2:3. When the same amount of heat is
supplied to both, the changes in their lengths are in the ratio. (If their linear coefficients are in the
ratio 5 : 6)

(A) 1 : 1
(B) 5 : 2
(C) 5 : 12
(D) 12 : 5

18) 1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the
system when thermal equilibrium is reached? latent heat of fusion of ice = 3.36 × 105 J/kg and latent
heat of vaporization of water = 2.26 × 106 J/kg.

(A) 2000 g water

(B) 665 g steam and 1335 g water
(C) 665 g ice and 1335 g water
(D) 2000 g steam

19) Two rods A and B of lengths ℓA and ℓB have coefficients of linear expansion αA : αB = 1 : 4. If the
difference between the lengths of the two rods is to be independent of temperature, ℓA : ℓB should be

(A) 1 : 16
(B) 1 : 4
(C) 1 : 2
(D) 4 : 1

20) A hole is made in a thin metal disc, as shown. When the temperature of metal is raised, then:- (r1
= radius of disc, r2 = radius of hole)

(A) r1 increases but r2 decreases

(B) r1 increases and r2 increases.
(C) r1 & r2 both decreases
(D) r1 decreases & r2 increases.

SECTION-II

1) In an experiment four quantities a, b, c and d are measured with percentage errors 1%, 2%, 3%

and 4% respectively. A quantity P is calculated as The maximum percentage error in P

is____ .

2) if x = a – b, find percentage error in x.

3) What is percentage error in volume of a sphere, when error in measuring its radius is 2% ?

4) The density of a material in CGS system is 8 g/cm3. In a system of unit in which unit of length is 5
cm and unit of mass is 20 g, the density of the material will be _____.

5) Resistance is given by R = V/i, where V = (100 5) volt and i = (10 0.2) ampere. What is the
percentage error in R ?

CHEMISTRY

SECTION-I

1) What is the empirical formula of vanadium oxide if 7.36 gm of the metal oxide contains 4.16 gm of
metal. (At mass of V = 52)

(A) V2O3
(B) VO
(C) V2O5
(D) V2O7
2) The minimum amount of O2 which is required to convert 100 amu of H2 gas completely into H2O
will be

(A) 800 amu

(B) 100 amu
(C) 640 amu
(D) 320 amu

3) 500 mL of a liquid (sp. gravity = 1.5) contains 5 NA molecules. What is the molecular mass of the
liquid.

(A) 150 amu

(B) 100 amu

(C)

(D) 750 amu

4) The minimum volume at NTP will be occupied by:

(A) 1 g H2
(B) 1 g He
(C) 1 g equimolar mixture of H2 & He
(D) 1 g O2

5) 6 gm nitrogen on successive reaction with different compounds gets finally converted into 30 gm
[Cr(NH3)xBr2] Value of x is [Atomic mass : Cr = 52, Br = 80]

(A) 1
(B) 2
(C) 3
(D) 4

6) If number of protons in X2- is 15. Find number of electrons in X2+.

(A) 13
(B) 14
(C) 15
(D) 16

7) A sample of calcium carbonate is 80% pure. 25g of this sample is treated with excess of HCl. How
much volume of CO2 will be obtained at NTP.

(A) 4.48 litre

(B) 5.6 litre
(C) 11.2 litre
(D) 2.24 litre
8) The volume of one mole of water at 277 K is 18 ml. One ml of water contains 20 drops. The
number of molecules in one drop of water will be

(A) 1.07 x 1021

(B) 1.67 x 1021
(C) 2.67 x 1021
(D) 1.67 x 1020

9) The average molar mass of chlorine is 35.5 g mol-1. The ratio of 35Cl to 37Cl in naturally occurring
chlorine is close to :

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

10) Number of atoms in the following samples of substances is the largest in : (At wt of I = 127g)

(A) 127.0 g of iodine

(B) 48.0 g of magnesium
(C) 71.0 g of chlorine
(D) 4.0 g of hydrogen

11)

Actual mass of He-atom is:-

(A) 4g
(B)
(C)
(D)

12) 1.12 g of KOH dissolved in water and formed 500 mL solution. Calculate molarity of solution
(atomic weight of K = 39).

(A) 22.4
(B)
(C) 0.4 M
(D) 0.04 M

13) The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of
the compound will be:

(A) C2H2
(B) C3H3
(C) C2H4O
(D) C6H6

14) At certain temperature two moles of element A combines with five moles of element B to
produce two mole of C. If atomicity of A and B is 2. The formula of compound C is :-

(A) AB3
(B) A2B5
(C) AB5
(D) A5B2

15) The relative density of a particular oil is 0.89. Calculate the density of that oil

(A)
(B) 890 kg/mm3
(C)
(D)

16) Which one of the following has maximum mass -

(A) 1.12 litre H2O at 1 atm and 4ºC

24
(B) 6 × 10 O3 molecules
(C) 0.5 kg CaCO3
(D) 8.96 litre of O2 at 1 atm and 0ºC

17)

The largest number of molecules is in

(A) 36 g H2O
(B) 54 g N2O5
(C) 46 g C2H5OH
(D) 28 g CO

18) Equal number of atoms are contained in one gram-atom of each element, and that the same
number of molecules is found in one gram -molecule of any compound. The terms gram atomic
weight and gram molecular weight are used to refer to a fixed number (Avogadro's number 6.022 ×
1023) of particle. The term mole stands for the amount of material which contains these number of
particles.
What will be the number of total electrons in
24 gm

(A) 12
(B) 10
(C) 12 NA
(D) 10 NA
19) Mass ratio of NH3 and CO2 respectively for maximum product formation as per reaction :
2NH3 + CO2 → NH2COONH4

(A)

(B)

(C)

(D)

20)

4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)

2NO (g) + O2 (g) → 2NO2 (g)
If 20 ml of NH3 is mixed with 100 ml of O2. Volume of O2 consumed at the completion of above
reactions is

(A) 20 ml
(B) 85 ml
(C) 35 ml
(D) 100 ml

SECTION-II

1) 16 g of SOx occupies 5.6 litre at STP assuming ideal gas nature, the value of x is (Molar volume of
S.T.P = 22.4 Lt)

2) Total moles of gases produced upon complete decomposition of 102 g NH3 as following are:
NH3(g) → N2(g) + H2(g)

3)

What amount (in gm) of CaO will be produced by 2 moles of Ca?

2Ca + O2 2CaO (Nearest integer)

4) A complex compound of iron has molar mass = 2800 and it contains 8% iron by weight. The
number of iron atoms in one formula unit of complex compound is: [Atomic weight of Fe = 56]

5) The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of
water is 1.15 g/mL. The molarity of this solution is (Nearest integer)

MATHEMATICS
 SECTION-I

1) is divisible by :

(A) 34 but not by 14

(B) Both 14 and 34
(C) Neither 14 nor 34
(D) 14 but not by 34

2) If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of is equal to _____.

(A) 11
(B) 13
(C) 15
(D) 17

3) The polynomials and

, when divided by (x – 4) leave the same remainder. The value of k is:

(A) 2
(B) 1
(C) 0
(D) –1

4) If then

(A) (–5, 0)
(B)
(C)
(D)

5) If , the
complete solution set of values of x is:

(A)
(B)
(C)
(D) None of these

6) The number of real solutions of the equation,

is:

(A) 2
(B) 3
(C) 1
(D) 4

7) For defined values of a, b, c expression

has the value equal to:

(A)

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

8) The value of [π] – [–e] is, where [.] denotes

greatest integer function:

(A) 5
(B) 6
(C) 7
(D) 8

9) The sum of the series:

is equal to

(A)

(B)

(C)

(D)

10) Consider the number N = 8 7 a 2 7 9 3 1 b, where b is a digit at unit's place and a is a digit at
ten lakh's place. Answer the following questions.
The greatest value of b for which N is divisible by 8 is:

(A) 0
(B) 2
(C) 4
(D) 6

11) The product of the real roots of the equation

 , is:

(A)

(B)

(C)

(D)

12) P(x) is a polynomial of degree 5 with leading

coefficient unity such that
P(1) = 1, P(2) = 4, P(3) = 9, P(4) = 16, P(5) = 25.
The constant term in the expression of P(x) is:

(A) 110
(B) –100
(C) –120
(D) 24

13) The sum of all the real roots of the equation

is:

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

14) Set of all real values of x satisfying the inequality

is

If [.] represents greatest integer function, then is equal to

(A) 5
(B) 7
(C) 8
(D) 9

15) Solution of is:

(A)

(B)

(C)
(D)

16) Number of real solutions of the equation

is:

(A) None
(B) Exactly 1
(C) Exactly 2
(D) 4

17) Let for which , then is equal to:

(A) 2
(B) 3
(C) 6
(D) 12

18) If , which of the

following yields ?

(A) pq

(B)

(C)

(D)

19) Number of positive integral solutions of the

equation

(A) 4
(B) 6
(C) 8
(D) 10

20) Let . If the value of the

expression can be expressed in the form where , (p + q) has the value equal

(A) 410
(B) 610
(C) 510
(D) 540
 SECTION-II

1) Let be the solution of the following equations . Then 2

is

2) The solution of is . Then value of k is

3) has complete solution set

then Value of a + b + c +d =

4) Let S be the set of all real roots of the equation, . Then number
of elements in 'S' is

5) The product of all positive real values of x satisfying the equation 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. D A A C C A D B D D C A A D B D B B D B

SECTION-II

Q. 21 22 23 24 25
A. 14 3 6 50 7

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. C A A D D A A B D D C D D B D A A D A C

SECTION-II

Q. 46 47 48 49 50
A. 2 12 112 4 2

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. A B B A B A B B B B B C A D C C A C B B

SECTION-II

Q. 71 72 73 74 75
A. 1 2 14 1 1
 SOLUTIONS

PHYSICS

1) [αx] = [βt] = 1.

2) Z =

3) v α gphq
using dimansional analysis :

p=q= (half)

4) v =
[at] = [v]
[t] = [c]

5) m ∝ Ea vb Fc
a = 1, b = –2, c = 0.

6)

7)
⇒ [B] = L1/2

Now,
M1L1T–2 = [A].L3/2
⇒ [A] = M1L–1/2T–2

Now,

⇒
8) F = KAa vb c

=
F=
=
So, F = A1 v2 1

F = Av2

9) x = at + bt2

m = (b) h2

10)
0
= [M L1T–1]

11)

12)

13) Dimensions of will be same as dimensions of pressure

14) Conceptual.

15)
Let
By comparing,
x=1

16)

Heat capacity = ms
mASA = mBSB
ρA (8V)SA = ρB (12V)SB

17)

18) Heat released by the steam when it converts to water at 100°C = 1×2.26×106 = 2260KJ
Heat given to ice to convert in water at 0°C = 1×3.36×105 = 336 KJ
Heat given to rise temp. 0° to 100°C
= 1 × 4.2 × 103 × 100 = 420 KJ
So for Heat required = Heat given
1504 KJ heat given by the system
2 kg water 100°C
↓
x Kg water
(2–x)kg steam
(2–x) × 2.26 × 106 = 1504 × 103
x = 1335 g water

19)
ℓA(1 + αAΔT) – ℓB(1 + αBΔT) = ℓA – ℓB
ℓAαAΔT – ℓBαBΔT = 0

ℓA : ℓB = 4 : 1
 20) Photographic expansion.

21)
= 3 × 1% + 2 × 2% + 3% + 4%
= 14%

22)

= 3%

23)

= 6%

24)

25)

CHEMISTRY

26) V O
mass 4.16 7.36 – 4.16

moles
2 5
∴ V2O5
27) O2 + 2H2 —→ 2H2O

25 molecule = 50 molecule
= 25 × 32 = 800 amu.

28) Mass of liquid is 750 gm and contains 5 moles.

29) 1 g O2 has minimum number of moles.

30)
On applying POAC on N atom

⇒ 212 = 53x

31)

The number of protons in x2- is 15.

The number of protons is equal to the number of electrons in a neutral atom.
Therefore, the number of electrons in neutral X is 15.
The number of electrons in X2+ is 13.

32) Purity concept:

- Given mass of reactant sample mass of pure reactant
- Mass of pure reactant sample = mass of pure reactant
+ mass of impurity

Purity calculation:
- Purity % = Given mass of pure reactant / mass of
reactant sample × 100

Given: 25 g sample

Pure CaCO3

Volume of CO2

33) Step 1:
Since 1 ml contains 20 drops and 1 mole of water is 18 ml, the number of drops in 1 mole is:
Step 2:
1 mole of water contains Avogadro's number of molecules:

Step 3:
The number of molecules in one drop of water is approximately .

34)

The average molar mass is given by:

step 1:

step 2:
step 3: The ratio of .

35)

4g of hydrogen = 4 mole of hydrogen

71.0 gm of chlorine = 1 = moles of chlorine

127 gm of iodine mole of I2

48 gm of magnesium

4.0 gm H2 has largest number of atoms.

36)

Thus 1 amu

Atomic mass of an element

Actual mass of an element = atomic mass (amu)

37)

38)

In C6H6 Total number of C = 12 6 = 72

Total number of H = 1 6 = 6
Molecular mass = 78.

39)

Ratio of volumes = A : B : C = 2 : 5 : 2
Ratio of no. of molecules in A, B and C = 2 : 5 : 2 or 1 : 5/2 : 1
1 molecule C contains 1 molecule (i.e. 2 atom) of A and 5/2 molecules (i.e. 5/2 × 2 atoms) of B
Hence the formula of C is A2B5

40)

Given that, Relative density

We know that, density of water at 4oC is = 1000 kgm-3

In CGS unit

41) (A) = ⇒ mass = 1.12 kg = 1120 gm

(B) 10 moles O3 ≡ 480 gm O3
(C) 500 gm CaCO3

(D)

42)

Number of molecules present in 36 g of water

Number of molecules present in 28 g of CO

Number of molecules present in 46 g of C2H5OH

Number of molecules present in 54 g of N2O5

Here, NA is Avogadro's number. Hence, 36 g of water contain the largest (2NA) number of
molecules.

43) mol
So moles of electrons = 1 × 10 and number of electrons = 10NA
44)

2NH3 + CO2 → NH2COONH4

45)

Volume of O2 used = 35 ml

46)

47)
6 mole After reaction 0 3 9
Total moles = 12

48) step 1:
From the balanced equation, 2 moles of Ca produce 2 moles of CaO.
Therefore, 2 moles of Ca will produce:

step 2:
Multiply the moles of CaO by its molar mass:

2 moles of Ca will produce 112.16 g of CaO

49) Percentage mass of an Fe

8= ⇒x=4

50)

Given density of solution = 1.15 g/mL

Mass of solution = 1000+120 = 1120 gm
Molar mass = 60
Volume = mass / density of solution
= 1120/1.15
No. of moles = 120/60 = 2
Molarity = No. of moles/volume
 = 2.05 M

MATHEMATICS

51)

is divisible by 6

also us divisible by 17
Given expression is divisible by 34 but not by 14.

52)

a2 + b2 + c2 = (a + b + c)2 – = –3

= 9 – 2(–2) = 13

53)

P(x) = Kx3 + 3x2 – 3

and Q(x) = 2x3 – 5x + k
Div. x – 4 and remainder are same
So P(4) = Q(4)
64 k + 48 – 3 = 2 × (4)3 – 5 × 4 + k
64 k + 48 – 3 = 128 – 20 + k
63 k = 108 – 45
k=1

54)

55)
56)

|x|2 – |x| – 12 = 0
(|x| + 3)(|x| – 4) = 0
|x| = 4 ⇒ x = ±2

57)

58)
[3.14] – [–2.7]
(3) – (–3) = 6

59)

60)

N = 8 7 a 2, 7 9, 3 1 b
Divisibility rule of 8: last three digits divisible by 8
312 is divisible by 8

61)

9x2 – 18|x| + 5 = 0
9|x|2 – 15|x| – 3|x| + 5 = 0 ( x2 = |x|2)
3|x| (3|x| – 5) – (3|x| – 5) = 0

Product of roots

62)

P(x) - polynomial of degree 5

P(1) = 1, P(2) = 4, P(3) = 9, P(4) = 16, P(5) = 25

Let
Implies x = 1, 2, 3, 4, 5 are the roots of q(x)

Constant term
= –120

63)

For

For
x=4
Hence sum

64)
Let logex = a
4a3 – 8a2 – 11a + 15 0

65)

For

For

for

66)
so option C is correct.

67)

68)

69)

Given

By (1) and (2)

hk = 32
(h, k) = (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1)
Solutions =
Number of solutions = 6

70)
Similarly

xy = 2

71)

72)
Case 1 :

Case 2: x < 0
Let

But

k=2

73)
For

a = 2, b = 3, c = 4, d = 5
a+b+c+d=2+3+4+5
= 14

74) Let 3x = t ; t > 0

t(t – 1) + 2 = |t – 1| + |t – 2|
t2 – t + 2 = |t – 1| + |t – 2|
Case-I : t < 1
t2 – t + 2 = 1 – t + 2 – t
t2 + 2 = 3 – t
t2 + t – 1 = 0
 is only acceptable
Case-II : 1 t<2

D < 0 no real solution

Case – III:
t2 – t + 2 = t – 1 + t – 2
t2 – 3t 5 = 0 D < 0 no real solution

75)
Take both side

Let

Product: