08-06-2025

6001CJA10100225001 JM

PHYSICS

SECTION-I

1) Given where F is force, x is distance and t is time. The dimensions of is :-

(A) MLT–1
(B) ML–1T
(C) M–1LT
(D) ML2T–1

2) You measure two quantities as

We should represent B–A as

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

3) The approximate value of x where x = sin 1° cos 1°, is

(A)

(B) 2

(C)

(D)

4) y = then is :

(A) +

(B)
–

(C)
–
(D) +

5)

(A)

(B) 1
(C) –1

(D)
–

6) Figure shows graph and slope relationship correct match :

List-I List-II

(P) (1) Positive

(Q) (2) Negative

(R) (3) Zero

(S) (4) can not determine

(A) P → 2 ; Q → 1 ; R → 3 ; S → 1
(B) P → 1 ; Q → 3 ; R → 1 ; S → 2
(C) P → 2 ; Q → 3 ; R → 1 ; S → 1
(D) P → 1 ; Q → 2 ; R → 3 ; S → 1

7) Which of the following equation is the best representation of the given graph ?

(A) y = 2x2
(B) x = 2y2
(C) y = –2x2
(D) x = –2y2

8) If the volume of a sphere increases at constant rate . If radius of the sphere is denoted
by r, the surface area of the sphere increases at the rate

(A)

(B)

(C)

(D)

9) The given graph represents x = y2. Then the area of shaded region is :

(A)
m2

(B)
m2

(C)
m2

(D)
m2
10) A wire has a mass , radius and length . The maximum
percentage error in the measurement of its density is:

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

11)

The radius of a circle is stated as 2.12cm. Its area should be written as :

(A) 14 cm2
(B) 14.1 cm2
(C) 14.11 cm2
(D) 14.1124 cm2

12) The vernier constant of a vernier callipers is 0.001 cm. If 49 main scale divisions coincide with
50 vernier scale divisions, then the value of 1 main scale division is :

(A) 0.1 mm
(B) 0.5 mm
(C) 0.4 mm
(D) 1 mm

13) The pitch of a screw gauge is 0.5 mm and there are 100 divisions on it circular scale. The
instrument reads + 2 divisions when nothing is put in-between its jaws. In measuring the diameter
of a wire, there are 8 divisions on the main scale and 83 division coincides with the reference line.
Then the diameter of the wire is

(A) 4.05 mm
(B) 4.405 mm
(C) 3.05 mm
(D) 1.25 mm

14) Given below are two statements : one is labelled as Assertion(A) and the other is labelled as
Reason (R).
Assertion (A) : In Vernier calliper if positive zero error exists, then while taking measurements, the
reading taken will be more than the actual reading.
Reason (R) : The zero error in Vernier Calliper might have happened due to manufacturing defect
or due to rough handling.
In the light of the above statements, choose the correct answer from the options given below :

(A) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(B) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(C) (A) is true but (R) is false
(D) (A) is false but (R) is true
15) A child loves to watch as you fill a transparent plastic bottle with shampoo. Horizontal cross-
sections of the bottle are circles with varying diameters. You pour in shampoo with constant volume
flow rate 16.5 cm3/s. At what rate is its level in the bottle rising at a point where diameter of the
bottle is 6.30 cm. Calculate keeping significant figures in view.

(A) 0.063 cm/s

(B) 0.63/cm/s
(C) 0.529 cm/s
(D) 0.5291 cm/s

16) The number of divisions on the circular scale is 100 and the screw moved by 1 mm when two full
rotations are given. Least count of the screw gauge is :

(A) 0.01 mm
(B) 5 × 10–2 mm
(C) 5 × 10–3 mm
(D) 1 × 10–3 mm

17) In ordinary Vernier calipers, 10th division of the Vernier scale coincides with 9th division of the
main scale. In a specially designed Vernier calipers the Vernier scale is so constructed that 10th
division on it coincides with 11th division on the main scale. Each division on the main scale equals
to 1 mm. The calipers have a zero error as shown in the figure-I. When the Vernier caliper is used to
measure a length, the concerned portion of its scale is shown in figure-II.

(A) Zero error in the calipers has magnitude 0.5 mm

(B) The length being measured is 1.08 cm
(C) The length being measured is 1.22 cm
Though the given Vernier scale does not follow the principle of Vernier scale, yet can be used
(D)
satisfactorily.

18) If x = a cos t, y = b sin t then equals :

(A)
(B)

(C)

(D)

19) Value of is :

(A)

(B)
2
(C) x + logex + 2x + C

(D)

20) Two bugs, one in blue colour and another in black colour. Start walking along x-axis the figure
shows their position on x-axis with time. At what time is the distance between them is minimum.

(A) Only at 1 sec

(B) At 1 sec and 5 sec
(C) At 1 sec, 3 sec and 5 sec
(D) only at 5 sec

SECTION-II

1) The value of Stefan’s constant in CGS system is . its value in SI

unit is given by . Find n.

2) The measurement of lengths of sides of a rectangle are

The maximum error in the area of rectangle is given by___ m2.

3)
The specific resistance ρ of a circular wire of radius r, resistance R, and length is given by

. Given :

r = 2.4 ± 0.02 cm, R = 30 ± 1 Ω, and

= 4.80 ± 0.01 cm. The percentage error in ρ is nearly (integer)

4) A physical quantity P is related to three observables A, B and C as . The percentage

error of the measurement in A, B and C are 1%, 0.5% and 2% respectively. Determine % error in the
quantity P.

5) In a new system of units, unit of mass is 2 kg, unit of length is 4 m and unit of time is 2 second.
How much joule is equal to 1 unit of energy in this new system?

CHEMISTRY

SECTION-I

1) Two isotopes of an element are present in sample in 2 : 3 mol ratio of mass number (M + 0.5) and
(M + 1) respectively. The mean atomic mass of element will be :

(A) 5M + 4
(B) 5M + 0.4
(C) M + 0.8
(D) M + 0.5

2) How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at NTP.

KClO3 KCl + O2

(A)
mol

(B)
mol

(C)
mol

(D)
mol

3) If the mass of proton is doubled and that of neutron is halved, the molecular weight of CO2,
consisting only C12 and O16 atoms, will

(A) Not change

(B) Increase by 25%
(C) Decrease by 25%
(D) Increase by 50%

4) Find the ratio of the number of atoms present in 16 g of O2 and 32 g of O3

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

5) The vapour density of a mixture containing NO2 and N2O4 is 27.6. The mole fraction of N2O4 in the
mixture is:

(A) 0.1
(B) 0.2
(C) 0.5
(D) 0.8

6) Calculate density of a gaseous mixture which consist of 3.01 × 1024 molecules of N2 and 32 g of O2
gas at 3 atm pressure and 860 K temperature (Given : R = 1/12 atm L/mole K)

(A) 0.6 g/L

(B) 1.2 g/L
(C) 0.3 g/L
(D) 12 g/L

7) 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

8) Total number of electrons present in 4.5 g oxalic acid (H2C2O4) is:

(Given : Molecular wt. of H2C2O4 = 90)
(Atomic Number : H = 1, C = 6, O = 8)

(A) 0.05 NA
(B) 2.3 NA
(C) 2.2 NA
(D) 2.1 NA

9) The percentage by mass of hydrogen in water is :

(A) 1.11%
(B) 11.11%
(C) 8.89 %
(D) 88.9 %

10) Which of the following species is isotonic with ?

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

11) An organic compound on analysis gave C = 54.4%, H = 9.1% and rest oxygen by mass. Its
empirical formula is :

(A) CHO2
(B) CH2O
(C) C2H4O
(D) C3H4O

12) 3 m NaOH solution has a density of 1.10 g/ml. The molarity of the solution is :-
(Molecular wt. of NaOH = 40)

(A) 2.94
(B) 3.25
(C) 3.64
(D) 1.25

13) What is the mass of H2O in 3M NaOH (40 g/mole) having volume = 500 ml and density = 1.2
g/ml.

(A) 540 g
(B) 480 g
(C) 600 g
(D) None of these

14) With increase of temperature, which of these changes?

(A) Molality
(B) Weight fraction of solute
(C) Molarity
(D) Mole fraction

15) An element exist in two isotopic forms X120 and X122, in equal abundance. The average atomic
mass of element is 121.44. If each atom of X120 are 10.04 times heavier than one C12 atom, then how
many times each atom of X122 is heavier than one C12 atom ?

(A) 10.04
(B) 10.4
(C) 10.2
(D) 10.167

16) The volume of Cl2 at STP react with 1 gm Fe to produce FeCl3 will be (Atomic mass : Fe = 56)

(A) 0.6 litre

(B) 1.2 litre
(C) 1.8 litre
(D) 2.2 litre

17) In a given reaction, 9 g of Al will react with how many grams of

(A) 6g
(B) 8g
(C) 9g
(D) 4g

18) Which of the following solutions will have maximum amount of NaOH?

(A) 4L of 0.1M NaOH soln

(B) 2L of 5% (w/v) NaOH soln
(C) 540g of 10% (w/w) NaOH soln
n
(D) 20 mole NaOH sol having mole fraction of H2O = 0.9

19) If 0.5 mol of CaBr2 is mixed with 0.2 mol of K3PO4, the maximum number of moles of Ca3(PO4) 2
that can be formed is:

(A) 0.1
(B) 0.2
(C) 0.5
(D) 0.7

20) 5 mole mixture of KClO3 & CaCO3 is heated strongly. If volume of total gas produced at NTP is
134.4 L. The mole precentage of CaCO3 in the mixture will be -

(A) 40%
(B) 60%
(C) 66.67%
(D) 80%

SECTION-II

1) The haemoglobin from the red corpuscles of most mammals contains about 0.33% iron. Physical
measurement indicate that hemoglobin is a very large molecule with a molar mass of about
How many moles of Fe are there in one mole of haemoglobin? [Fe = 56 u]

2) 16 g of SOx gas occupies 5.6 L at 1 atm and 273K. What will be the value of x ?

3) 50 % pure sample of CaCO3 heated with 80 % yield to produce 0.448 litre CO2 at 1 atm and 273
K temperature. Calculate mass of CaCO3 sample in gm.

4) A complex compound of iron has molar mass = 2800 g/mol & it contain 8% iron by weight. The
number of iron atoms in one formula unit of complex compound is:
[At. wt. Fe = 56]

5) A gas is found to have the formula (C3O2)n. Its vapour density is 34. The value of n will be ......
(At. wt of C = 12, O = 16)

MATHEMATICS

SECTION-I

1) Ten times the number is

(A)

(B)

(C)

(D)

2) If x2 + y2 + 4z2 – 6x – 2y – 4z + 11 = 0, then xyz is equal to

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

3) If is divisible by (x-2)2, then is equal to

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

(D)
4) The number of real solutions of the equation
|x|2 – 3|x|+ 2 = 0 is

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

5) The solution set of is :

(A) [1, 5]

(B)

(C)

(D)

6) The solution of is

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

7) Number of values of x which is satisfy the equation log3(3x – 8) = (2 – x)

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

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

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

9) A polynomial is exactly divisible by x + 1 and when it is divided by 3x – 1 the remainder is 4, find

the remainder when it is divided by
3x2 + 2x – 1.

(A) 2x + 3
(B) 3x + 2
(C) x + 4
(D) 3x + 3

10) simplifies to :

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

11) If x + y = 1 and x2 + y2 = 2 then the value of

x4 + y4 equals

(A) 7
(B) 6

(C)

(D)

12) When 12 is divided by a positive integer n, the remainder is (n – 3), which of the following is the
possible value of 'n' ?

(A) 7
(B) 8
(C) 11
(D) 5

13) Number of real roots of equation:

is

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

14) The exhaustive set of x for which is

(A) (–5, –4) ∪ (–3, 1)

(B) (–∞, –5) ∪ (–4, –3] ∪ [1, 3) ∪ (3, ∞)
(C) (–∞, –5) ∪ (4, 3) ∪ (3, +∞)
(D) [1, ∞)

15) The sum of the real values of x satisfying the equation |x2 + 4x + 3| + 2x + 5 = 0 is
(A)
(B)
(C)
(D)

16) The value of is :

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

17) The value of [e2] – [– e2] is, where [.] denotes greatest integer function -

(A) 19
(B) 15
(C) 10
(D) 18

18) The polynomials P(x) = kx3 + 3x2 – 3 and

Q(x) = 2x3 – 5x + k, when divided by (x – 4) leave the same remainder. The value of k is

(A) 2
(B) 1
(C) 0
(D) –1

19) If a > b > 0 are two real numbers, the value of ,

is :

(A) independent of b
(B) independent of a
(C) independent of both a & b
(D) dependent on both a & b.

20) If then x3 – 3x is equal to

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

SECTION-II
1) If x = , then [x] = ?, where
[ ] represent greatest integer function :

2) |x – 2| + |x – 3| – |x – 4| = 10, the number of solution of the equation will be :

3) The smallest positive integer which satisfy the inequality 2x4 – 3x3 > 9x2 is :

4) The value of x satisfying the equation , is :

5) If and and , where

m and n are relative primes, the value of 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 B C B B C B B A D B B B B C C C B A B

SECTION-II

Q. 21 22 23 24 25
A. 8 6 5 6 8

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

SECTION-II

Q. 46 47 48 49 50
A. 4 2 5 4 1

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

SECTION-II

Q. 71 72 73 74 75
A. 2 2 4 25 10
 SOLUTIONS

PHYSICS

1)

2)

3)

4)

5)

6) Slope → It is tangent of angle made by any line with +ve x-axis measured in anticlockwise
sense.

7)

The curve is best represented by a parabola symmetric about x-axis

k is any positive constant.

8)
9) A =

A=

=
A = 12

10)

% Error in P = %M +2%r + %ℓ

= (.01 + .02 + .01) × 100 ⇒ 4%

11)

Round off to 3 significant figure.

12)

0.5 mm

13)

4.405 mm

14) Assertion & Reason both are correct

Theory

15)

cm/s

16)

5 × 10–3 mm
 17) VSD = 1.1 mm and 1 MSD = 1.0 mm
Least count = LC = MSD ~ VSD = 0.1 mm
For (A) : Zero error = – (No. of divisions of vernier scale coincident) × LC = –7 × 0.1 = – 0.7
mm
For (C) : Reading = (No. of divisions coincident on main scale) × MSD – (No. of divisions
coincident on vernier scale ) LC– zero error = 1.22 cm

18)

19)

20) The minimum distance between the bugs is 0.

21)

22)

Area A = ab

23)

24)

25)

CHEMISTRY

26)
=
Mave = M + 0.8

27) KClO3 KCl + O2

mole or 33.6 litre O2 from 1 mole KClO3

11.2 litre of O2 formed by mole KClO3

28) Old Molecular weight = 12 + 32 = 44

New Molecular weight = 15 + 40 = 55

29)

30)

Vapour density of mixture = 27.2

Mavg = 2 × VD
= 2 × 27.2 = 55.2
Let NO2 N2O4
x% 100–x%

N2O4 = 100 – 80% = 20% = 0.20.

31)
Moles of N2 = = 5 mole

Moles ofO2 = = 1 mole

PMavg = dmix RT

dmix = 1.2 g/L

32) 1 g O2 has minimum number of moles.

33)

34) 18 g of H2O contains 2g H2

∴ 100 g of H2O = = 11.11%

35) Isotonic ⇒ Same number of neutrons

Atom Number of Neutrons
8
6
7
8
10

36)

37)

M = 2.94
38) Total mass = 600 g

Mass NaOH =

39)

Ans : (C)
Molarity of solution depends upon temp

40) Mass of X120 atom = 10.04 × Mass of C12 atom

Mass of X120 atom = 12 × 10.04 = 120.48

Avg. atomic mass = (x ⇒ Atomic mass of X122)

x = 122.4

The number of times X122 is heavier than C12 is

41)

moles of

mole of
Requirement volume of Cl2 at STP

42)

Weight of = 8g

43) 2 L of 5% (w/v) NaOH solution contains 100 g NaOH

44)

According to the chemical reaction.

2 mol of K3PO4 reacts with 3 mol of CaBr2
0.2 mol. of K3PO4 reacts with 0.3 mol of CaBr2
Moles of CaBr2 left unreacted = 0.5 – 0.3 = 0.2
Hence, K3PO4 is the limiting reagent, since it reacts completely and it decides the amount of
products formed
2 mol of K3PO4 gives = 1 mol of Ca3(PO4)2
0.2 mol of K3PO4 gives = 0.1 mol of Ca3(PO4)2
Thus, the answer is (A).

45)

So, 1.5 x + 5-x =

Hence, x = 2 mol

% CaCO3 = ×100 = 60%

46)
n × 56 = 33 × 6.8
⇒n=4

47)
M = 64
64 = 32 +16x
x=2

48) CaCO3(s) → CaO(s) + CO2 (g)

No. of mole of CO2 = = 2 × 10–2

0.8n = 2 × 10–2

⇒n= = 0.025
= 2.5 gm (pure)
Mass of CaCO3 sample = 5 gm.

49) 8% of Molar mass = n × 56

50) Molar mass of the gas (C3O2) = n (12 × 3 + 16 × 2) = n × 68

∴ n × 68 = 2 × Vapour density
n × 68 = 2 × 34
n=1
MATHEMATICS

51)

52) x2 + y2 + 4z2 – 6x – 2y – 4z + 11 = 0
⇒ (x–3)2 + (y–1)2 + (2z–1)2 = 0

⇒ x = 3 , y = 1, z =

⇒ xyz =

53) is divisible by (x-2)

Let
again Q(x) is divisible by x - 2 and by factor theorem

54) |x|2 – 3|x| + 2 = 0

Case – I
x>0
x2 – 3x + 2 = 0
(x–2)(x–1) = 0
x = 1,2

Case – II
x<0
x2 + 3x + 2 = 0
(x + 2)(x + 1) = 0
x = –1, –2
Number of real solution = 4

55)
56)

Critical points : |x-1|-2=0 |x-1|=2

x-1=2 or -2
x=3 or -1

case:1 -
||x-1|-2|=x-3 |x-1-2|=x-3
|x-3| = x-3 true for all
case:2 -
|x-1-2| = 3-x |x-3|=3-x
true for
case:3 -
|1-x-2|=3-x |-x-1| = 3-x
|x+1=3-x
x+1=3-x or x-3 rejected
x=1 rejected
from case 1 and case 2

57) log3(3x – 8) = 2 – x
3x – 8 > 0
3x – 8 = 32–x
3x – 8 = 32.3–x
Let 3x = t

t–8=
t2 – 8t – 9 = 0
(t – 9)(t + 1) = 0
t = 9, t = –1
3x = 9, 3x = –1 rejected
x=2
3x – 8 > 0
32 – 8 > 0
9–8>1
Ans. x = 2

58)

59)
since, P(x) divisible by

divisible by 3x-1then remainder is 4;

solving (i) & (ii), a=3, b=3

so remainder = 3x + 3

60)

=
=

61)

x + y = 1 x 2 + y2 = 2
(x+y)2 = x2 + y2 + 2xy = 1
2+2xy = 1
2xy = -1 xy = -1/2
x + y = (x + y2)2 - 2x2y2
4 4 2

= 22 - 2x (-1/2)2

= 4 - 2× = 4

62) check from option

A. If n = 7, remainder = 5

B. If remainder = 4

C. If n = 11, remainder = 1

d. If n = 5 remainder = 2
n-3 = 2
hence (D) correct

63)
64) x = 1, - 3, 2 (zeros) and x = -4, 3, -5 points where denominator is zero.

65) If x2 + 4x + 3 = (x + 3)(x + 1) ≥ 0,
x ∈ R – (–3, – 1) ... (1)
The given equation becomes,
x2 + 6x + 8 = 0 ⇒ x = –2, –4 ... (2)
from (1), (2) ⇒ x = –4
If x2 + 4x + 3 < 0, x ∈ (–3, –1) ... (3)
2
The equation becomes, –(x + 4x + 3) + 2x + 5 = 0
or x2 + 2x – 2 = 0 ⇒ ... (4)
from (3), (4) ⇒

Sum of the roots

66)
=

67)

[e2] – [– e2]
=[7.3]-[-7.3]
=7+8=15

68)

P(4) = 64k + 48 – 3 = 64k + 45

Q(4) = 128 – 20 + k = k + 108
Given P(4) = Q(4)
∴ 64k + 45 = k + 108
⇒ 63k = 63 ⇒ k = 1

69)
⇒ (y –a) (y +b) = 0
⇒ y =a
rejected

70) Take cube both sides

71)

72)

|x – 2| + |x – 3| – |x – 4| = 10

Case-I:

Case-II:

rejected
Case-III:

rejected
Case-IV:

73) x2 (2x2 – 3x – 9) > 0

x2 (2x + 3) (x – 3) > 0

Ans = 4

74) Using repeatedly, we get ⇒

⇒ ⇒ x = 25.

75)
From Eqs. (i) and (ii), we get