02-08-2025

1001CJM202036250002 JM

PART-1 : PHYSICS

SECTION-I

1) A radioactive nucleus undergoes a series of decay according to the scheme

If the mass number and atomic number of A are 180 and 72

respectively, what are these number for A4?

(A) 172, 68
(B) 170, 69
(C) 172, 69
(D) 168, 68

2) If particle of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V, the de-
broglie wavelength found is λ. Now if mass is doubled keeping the charge same and accelerated by a
potential difference of 2500 V, then the de-broglie wave length found is:

(A) 0.1 λ
(B)
(C) 10λ
(D)

3) Fission of one nucleus 92U235 releases 250 MeV of energy. The number of fissions per second
required to produce 1 MW power is

(A) 9.2 × 1017

(B) 6.3 × 1023
(C) 1.6 × 1019
(D) 2.5 × 1016

4) A particle of mass M at rest decays into two particles of masses m1 and m2 having non-zero
velocities. The ratio of the de-Broglie wavelengths of the particles λ1 : λ2 is :-

(A)

(B)

(C) 1 : 1

(D)
5) A nuclear reactor delivers a power of 109 W. What is the amount of fuel consumed by the reactor
in one hour ?

(A) 0.04 g
(B) 0.08 g
(C) 0.72 g
(D) 0.96 g

6) The de-Broglie wavelength of a particle moving with a velocity 2.25 × 108 m/s is equal to the
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is
(velocity of light is 3 × 108 m/s)

(A) 1/8
(B) 3/8
(C) 5/8
(D) 7/8

7) The wavelength of de-Broglie wave is 2μm, then its momentum is (h = 6.63 × 10–34 J.S)

(A) 3.315 × 10–28 kg - m/s

(B) 1.66 × 10–28 kg - m/s
(C) 4.97 × 10–28 kg - m/s
(D) 9.9 × 10–28 kg-m/s

8) Kinetic energy of the emitted α-particle in the

α-decay of will be,(where mα = 4.00260u,
= 226.02540 u
and = 222.01750 u ):-

(A) 8.93 MeV

(B) 4.93 MeV
(C) 8.77 MeV
(D) 4.85 MeV

9) An electron of mass ‘m’ and charge ‘e’ initially at rest gets accelerated by a constant electric field
E. The rate of change of de-Broglie wavelength of this electron at time t; ignoring relativistic effect
is :-

(A)

(B)

(C)

(D)
10) The potential energy of a particle of mass m is given by,

λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0 ≤ x ≤ 1 and x > 1 respectively. If the
total energy of the particle is 2E0, the ratio λ1 / λ2 will be :-

(A) 1
(B) 2
(C)

(D)

11) Two identical parallel plate air filled capacitors are placed in series and connected to a constant
voltage source of V0 volt. If one of the capacitors is completely immersed in a liquid with dielectric
constant K, the potential difference between the plates of the other capacitor will change to :-

(A)

(B)

(C)

(D)

12) Three capacitors are connected as shown in fig. Then the charge on capacitor C1 is :-

(A) 6μC
(B) 12μC
(C) 18μC
(D) 24μC

13) A parallel plate capacitor is charged to a potential difference of 50 V. It is discharged through a

resistance. After 1 second, the potential difference between plates becomes 40 V. Then incorrect
option is :-

(A) Fraction of stored energy after 1 second is 16/25

(B) Potential difference between the plates after 2 seconds will be 32 V
(C) Potential difference betweeen the plates after 2 seconds will be 20 V
 Time constant of discharging circuit is
(D)

14) Three capacitor A, B and C are connected to a battery of 25 volt as shown in the figure. The ratio

of charges on capacitors A, B and C will be :-

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

15) As the switch 'S' is closed in the circuit shown in figure, current passed through it is :-

(A) 4.5 A
(B) 6.0 A
(C) 3.0 A
(D) Zero

16) The resistance in the two arms of a meter bridge are 5 Ω and R Ω, respectively and galvanometer
shows zero deflection. When the resistance R is shunted with an equal resistance, the new balance

point is at 1.6 ℓ1. The resistance ‘R’ is :-

(A) 10 Ω
(B) 15 Ω
(C) 20 Ω
(D) 25 Ω

17) Power generated across a uniform wire connected across a supply is H. If the wire is cut into n
equal parts and all the pars are connected in parallel across the same supply voltage, the total
power generated in the wire is :-

(A)

(B) n2H
(C) nH

(D)

18) When a current of 4 A flows within a battery from its positive to negative terminal, the potential
difference across the battery is 12 volts. The potential difference across the battery is 9 volts when a
current of 2 A flows within it from its negative to its positive terminal. The internal resistance and
the e.m.f. of the battery are :-

(A) 0.1 Ω, 4V
(B) 0.2 Ω, 5V
(C) 0.5 Ω, 10V
(D) 0.7 Ω, 10V

19) A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations

(i) key K is kept closed and plates of capacitors are moved apart using insulating handle
(ii) key K is opened and plates of capacitors are moved apart using insulating handle.
Which of the following statements is correct?

(A) In (i), Q remains same but C changes.

(B) In (ii) V remains same but C changes.
(C) In (i) V remains same and hence Q changes.
(D) In (ii) both Q and V changes.

20) All straight wires are very long. Both AB and CD arc of the same circle, both subtending right

angles at the centre O. Then the magnetic field at O is :-

(A)
(B)

(C)

(D)

SECTION-II

1) A nuclear reactor generates power by fission of 92U235 into two equal fragments of 46Pd116 with the
emission of two γ-rays of 5.2 MeV each and three neutrons. The average binding energies per
nucleon of 92U235 and 46Pd116 are 7.2 MeV and 8.2 MeV respectively. The usable energy released per
fission event is:

2) The binding energies per nucleon of deuteron (1H2) and helium (2He4) atoms are 1.1 MeV and 7

MeV. If two deuteron atoms react to form a single helium atom, then the energy released is MeV,
than x is :-

3) For what value of R in circuit, current through 4Ω resistance is zero.

4) A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω

resistance, it can be converted into a voltmeter of range 0–30 V. If connected to a resistance, it

becomes an ammeter of range 0 - 1.5 A. The value of n is :-

5) In the following figure, the charge on each condenser in the steady state will be :- (in µC)

PART-2 : CHEMISTRY
 SECTION-I

1) The density of 3M solution of Na2S2O3 is 1.25 g/ml. Then mole fraction of Na2S2O3 is :

(A) 0.65
(B) 0.5
(C) 0.065
(D) 0.37

2) Mole fraction of HNO3 in water is 0.5. What is the molality of HNO3 ?

(A) 18 m
(B) 1.8 m
(C) 55.5 m
(D) 6.2 m

3) 1.5 L of x% (w/w) NaOH solution has molality 5m. If solution contains 1000 gm of solvent. then
find % strength x :

(A)

(B)

(C)

(D)

4) PtCl4·6H2O can exist as a hydrated complex. Its 1 molal aq. solution has depression in freezing
point of 3.72. Assume 100% ionisation and Kf(H2O) = 1.86°C mol–1 kg, then complex is :-

(A) [Pt(H2O)6]Cl4
(B) [Pt(H2O)4 Cl2]Cl2·2H2O
(C) [Pt(H2O)3 Cl3]Cl·3H2O
(D) [Pt(H2O)2 Cl4]·4H2O

5) Correct order of freezing point of given aqueous solution :

Solution-I : 0.1m KCl,
Solution-II : 0.2 m HA (α = 0.5)
Solution-III : 0.1m C6H12O6

(A) I < II < III

(B) III < II < I
(C) III < I < II
(D) II < I < III

6)
 List-I List-II

Propan-2-ol
(P) (1) Maximum boiling azeotrope
+ 2-methyl propanol

(Q) H2O + HNO3 (2) Raoult's law is obeyed

(R) CS2 + CH3COCH3 (3) Positive deviation from Raoult's law

Fractional distillation column gives azeotrope from

(S) H2O + HCOOH (4)
residual liquid

(A) P → 2;Q → 1,4;R → 3;S → 1,4

(B) P → 2;Q → 3;R → 1;S → 4
(C) P → 1;Q → 2;R → 3;S → 4
(D) P → 1, 4;Q → 3;R → 2;S → 1, 3

7) In a binary ideal liquid solution of A and B in which vapour pressure of pure component A and B
are 600 and 400 torr respectively at temperature 'T'. Pick the correct statement :

(A) mole fraction of A in vapour phase is always greater than mole fraction of B in vapour phase.
(B) mole fraction of A in vapour phase is always greater than mole fraction of A in liquid phase.
(C) vapour pressure of solution can not exceed 600 torr at temperature T.
(D) mole fraction of B in vapour phase is always greater than mole fraction of B in liquid phase.

8) A liquid freezes at 7ºC and boils at 77ºC. If kf and kb of this liquid are 5.6 and 2.5 K / molal
respectively, then ratio of molar latent heat of fusion to molar latent heat of vaporisation is -

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

9)

Osmotic pressure of 0.01M glucose solution at 27ºC is ____.

[R = 0.082 L-atm/K/mole]
(A) 0.246 atm
(B) 0.24 atm
(C) 2.46 atm
(D) 2.65 atm

0 0
10) Two liquids A and B have, PA and PB = 1 : 3 at a certain temperature. Assume A and B form an
ideal solution, and the ratio of mole of A to B in the liquid phase are 1 : 3, then mole fraction of A in
vapour phase in equilibrium with the solution -

(A) 0.2
(B) 0.10
(C) 0.01
(D) 0.4

11) If P0 and PS are the vapour pressure of solvent and its solution respectively. N1 and N2 are the
mole fraction of solvent and solute respectively then

(A)

(B) P0 – PS = P0N2
(C) PS = P0N2

(D)

12) The value of observed and calculated molecular weight of silver nitrate are 92.64 and 170
respectively. The degree of dissociation of silver nitrate is :

(A) 60%
(B) 83.5 %
(C) 46.7%
(D) 60.23%

13) A solution of x moles of sucrose in 100 grams of water freezes at –0.2°C. As ice separates the
freezing point goes down to 0.25°C. How many grams of ice would have separated?

(A) 18 grams
(B) 20 grams
(C) 25 grams
(D) 23 grams

14) A solution containing 1.23 g of Ca(NO3)2 in 10 g of water has a normal boiling point of
100.975ºC. The degree of dissociation of Ca(NO3)2 will be : (Kb, water = 0.52 k kg mol–1)
(Ca = 40; N = 14, O = 16)

(A) 0.25
(B) 0.33
(C) 0.5
(D) 0.75

15) Mixture(s) showing positive deviation from Raoult's law at 35ºC is(are)

(A) carbon tetrachloride + methanol

(B) chloroform + acetone
(C) benzene + toluene
(D) phenol + aniline

16) A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose
(molar mass = 180 g mol–1) in 100 mℓ of water at 27ºC. The osmotic pressure of the solution is:
(R = 0.08206 L atm K–1 mol–1)

(A) 4.92 atm

(B) 1.64 atm
(C) 2.46 atm
(D) 8.2 atm

17) 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82°C.
If Na2SO4 is 81.5% ionised, the value of x.
(Kf for water = 1.86°C kg mol–1) is approximately.
(Molar mass of S = 32 g mol–1
and that of Na = 23 g mol–1)

(A) 45 g
(B) 65 g
(C) 15 g
(D) 25 g

18) 2NH3(g) N2(g) + 3H2(g) if degree of dissociation of ammonia at equilibrium is 0.7 then
observed molecular weight of reaction mixture at equilibrium -

(A) 12
(B) 10
(C) 15
(D) 17

19)

A(g) + 2B(s) 2C(g).

Initially 2 mol A(g), 4 mole of B(s) and 1 mole of a inert gas are present in a closed container. After
equilibrium has established, total pressure of container becomes 9 atm. If A(g) consume 50% at
equilibrium then calculate Kp for above reaction -
(A) 9 atm
(B) 36/5 atm
(C) 12 atm
(D) 6 atm

20) For the gas phase reaction SO3(g) ⇌ SO2 (g) + O2 (g)
if the gaseous mixture in a closed container is allowed to come at equilibrium and the degree of
dissociation (α) at equilibrium is found to be 2/3 at 400 K and 1 atm pressure then calculate the
vapour density of the equilibrium mixture.

(A) 30
(B) 60
(C) 40
(D) 20

SECTION-II

1) What volume (in ml) of 0.10 M H2SO4 must be added to 50 mL of a 0.10 M NaOH solution to make
a solution in which the molarity of the H2SO4 is 0.050 M?

2) 100 ml of 0.1 M HCl solution is mixed with 400 ml of 0.2 M HCl solution & is diluted by adding
500 ml water. Calculate final molarity of HCl solution.
[Express your answer is terms of 9 × 10–x & fill x in OMR sheet]

3) Find the molarity of 5.6% w/v KOH if density of solution is 1.4 g/ml :

4) If 1 gm of solute dimerise upto 75% in 100 gm of H2O & depression in freezing point 0.093°C.

Find molecular weight of solute .

Fill your answer as sum of digits (excluding decimal places) till you get the single digit
answer.

5) For the reaction 3 A (g) + B (g) 2 C (g) at a given temperature, Kc = 9.0. What must be the
volume (in litre) of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium?

PART-3 : MATHEMATICS

SECTION-I

1) The value of , where {.} denote fractional function -

(A) 0
(B) 1
(C) 1/2
(D) Does not exist

2) If , then which of the following is true :

(where [.] is greatest integer function)

(A) LHL exist but RHL does not exist

(B) LHL does not exist but RHL exist
(C) Both exist and equal
(D) Both exist but are not equal

3) The value of is

(A) 0
(B) 1
(C) 2
(D) Does not exist

4) is equal to :

(A)

(B) 1
(C) – π
(D) π

5) The value of is equal to -

(A) 0
(B) 1
(C) –1
(D) –2

6) is equals to

(A)

(B)

(C) 1
(D) 0

7)

The number of points where

ƒ(x) = (x2 – x) |x3 – 9x2 + 20x| is not differentiable, is-

(A) 2
(B) 3
(C) 0
(D) 4
8) If f(x) = max. (sin x, sin–1 (cos x)), then

(A) f is differentiable everywhere

(B) f is continuous everywhere but not differentiable at infinite number of points

(C)
f is non-differentiable at x = ,n∈I
(D) f(x) is continuous but non-differentiable at finite points

9) If , x ∈ [–2,2], then- (where [.] denotes greatest integer function)

(A) ƒ(x) is discontinuous at 4 points and not differentiable at 4 points

(B) ƒ(x) is discontinuous at 3 points and not differentiable at 4 points
(C) ƒ(x) is discontinuous at 4 points and not differentiable at 5 points
(D) ƒ(x) is discontinuous at 5 points and not differentiable at 5 points

10) If is continuous and differentiable at x = 1, then -

(A) a = –2 = –b
(B) a = 2 = –b
(C) a = 1 = b – 4
(D) a = –1 = b + 4

11) Let .
If ƒ(x) is differentiable in [0,2], then value of p is-

(A)

(B)

(C)

(D)

12) Number of points in [–2π,2π]

where ƒ(x) = |cos–1(cosx)| is non derivable is -

(A) 0
(B) 2
(C) 3
(D) 5
13) If is continuous ∀ x ∈ R
then which of the following holds good

(A) d = 2c
(B) a ≠ b
(C) a + b + d = –2

(D)

14) Find the value of p and q for which the function is differentiable
at x = –1

(A) p = 1, q = 3
(B) p = 2, q = 3
(C) p = 3, q = 2
(D) p = 3, q = 1

15) If , then ƒ'(–1) is

(A) can not be determined because ƒ(x) is not differentiable at x = –1

(B) 2
(C) 0
(D) 3

16) The value of ƒ(0) so that (where x > –1, x ≠ 0), is continuous is

(A)

(B)

(C)

(D) 3

17) If y = tan–1 , < x < π, then equals :-

(A) –1/2
(B) –1
(C) 1/2
(D) 1

18) If f(x) = x + tan x and f is inverse of g, then g'(x) is equal to :-

(A)

(B)

(C)

(D)

19)

(A)

(B)

(C)

(D)

20)

(A) (sin x)x (x cotx + ℓn sinx)

(B) (sin x)x (x cosx + ℓn sinx)
(C) (sin x)x (x cotx – ℓn sinx)
(D) (sin x)x (x cosx – ℓn sinx)

SECTION-II

1) is equal to

2)

The graph of the function y = f(x) is shown below.

If

then find the value of

[Note: [K] denotes greatest integer less than or equal to K]

3)

Let a, b ∈ R, b ≠ 0, Define a function

If f is continuous at x = 0, then 10 – ab is equal to _________.

4) ƒ is a differentiable and non zero function satisfies ƒ(x + y) = ƒ(x) + ƒ(y) + x2y + xy2, for all real

numbers x and y. If , then ƒ'(–2) is

5) If y = sec–1 + sin–1 then is equal to :-

 ANSWER KEYS

PART-1 : 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 A D C A B A D A C B A C A A B B C C C

SECTION-II

Q. 21 22 23 24 25
A. 200 236 1 5 12

PART-2 : 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 C B C D A B C A B B B B D A A A B A A

SECTION-II

Q. 46 47 48 49 50
A. 100 2 1 8 6

PART-3 : 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 A D B B A B C A A C D B B A A C A A

SECTION-II

Q. 71 72 73 74 75
A. 0 8 14 6 0
 SOLUTIONS

PART-1 : PHYSICS

1)
A – 8 = 180 – 8 = 172
Z – 3 = 72 – 3 = 69

2) Kinetic energy acquired by charge k = qv

3) Let number of fissions per seconds = n

∴ Energy per second = n × 250 MeV/s

∴ = n × 250 × 106 × 1.6 × 10–19 J/s

∴ Power = n × 4 × 10–11 W
∴ 1 MW = n × 4 × 10–11 W

or n = = 2.5 × 1016
16
∴ n = 2.5 × 10 .

4) [P1 = P2 = P]

5)

∴m= = =
–5
= 4 × 10 kg = 0.04 g

6) Kparticle =

⇒ ....(i)

Kphoton = .....(ii)
∴

7)
= 3.31 × 10–28 kg - m/sec

8)

mass defect =
= 226.02540 – [222.01750 – 4.00260]
= 0.0053 u
∴ Q value = (Δm)c2 = 0.0053 × 931 MeV
= 4.9343 MeV

and = 4.85 MeV

9) Acceleration produced due to electric field is

a=

Now, v = u + at =
de-Broglie wavelength of an electron is given by

λ=

10)

0
11) V1 = V ×
12)
2(6 – x) + 4 (0 – x) + 2 (6 – x) = 0
12 – 2x – 4x + 12 – 2x = 0
24 = 8x ∴ x = 3V
∴ qC1 = 2 (6 – 3) = 6 μC

13)
40 = 50 e–1/τ

e–1/τ = ...(1)
v = 50 e = 50(e–1/τ)2
–2/τ

= = 32 volt

= (e–1/τ)2

from (1)

e1/τ = ⇒

14) Let qA = q, qB = q=

qC =

∴ qA : qB : qC = q : : =5:2:3

15) Let V be the potential of the junction as shown in figure. Applying junction law, we have

or
or 40 – 2V + 5 – V = 2V
or 5V = 45 ⇒ V = 9V
∴

16)

ℓ1 = 25 cm

R = 15 Ω

17)

Initially H =

Power in one branch

Total power =

18)

19) When key K is kept closed, condenser C is charge to potential V. When plates of capacitor
are moved apart, its capacitance, C = decreases.
As potential of condenser remains same, charge Q = CV decreases. So option (3) is correct.
Once key K is closed, condenser gets charged, Q = CV
Now, if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains
same.

As plates of capacitor are moved apart, its capacity decreases

Therefore, its potential, V = increases.

20)

, B4 = 0, B6 = 0

Bnet =

21) 92U235 → 2x 46Pd116 + 20n1

B.E. 1692 1902.4
energy of two g - rays photons
= 2 × 5.2 = 10.4 MeV
∴ usable energy per fission
= (210.4 – 10.4) MeV
= 200 MeV

2 2 4
22) 1H + 1H → 2 He
BEPN 1.1 1.1 7
BE 2 × 1.1 2 × 1.1 4×7
ΔE = 28 – 4.4 = 23.6 MeV

23) 10–2i–Ri–4= 0
10–2i = 6 i=2
 24)
V = Ig(R + g)
30 = 0.006 (R + g)

R + G = 5000
⇒ G = 10Ω
IgG = (I – Ig)g
(.006)(10) = (1.494)(S)

S= ⇒∴n=5

25) In steady state current flows through 4Ω resistance only and is i = .

Potential difference across 4Ω resistance is V = 2 × 4 = 8 volt
Hence, potential difference across each capacitor is 4V
So charge on each capacitor Q = 3 × 4 = 12μC.

PART-2 : CHEMISTRY

26) W = 1000 ml × 1.25 g/ml = 1250 gm

Wsolute = 3×158 = 474 gm
Wsolvent = 1250 – 474 = 776 gm

nsolvent =

xsolute =

27)
28) 5m NaOH solution = 5 × 40 gm NaOH
Total Mass of solvent + solute = 200 + 1000 = 1200 gm

% NaOH =

29) ΔTf = i.kg. m

3.72 = i × 1.86 × 1
i=2
∵ 1 + α (n – 1) = 2
1+n–1=2
n=2

30)

[I] 0.1 m KCl [strong electrolyte]

α = 1 KCl → K+ + Cl–
n=2
⇒ i = 1 + α (n–1) = 1 + 1 (2–1)
⇒i=2
⇒ (ΔTf)I = i×m×kf = 2 × 0.1 kf = 0.2 kf ⇒ (Tf)I = –0.2 kf
(II) 0.1 m HA (a = 0.5)
HA H+ + A–
n=2
⇒ i = 1 + α (n–1)
⇒ i = 1+ 0.5 (2–1)
⇒ i = 1.5
⇒ (ΔTf)II = imkf = 1.5 × 0.2 kf = 0.3 kf ⇒(Tf)II = –0.3 kf
(III) 0.1 m C6H12O6 (not dissociated)
⇒i=1
⇒ (ΔTf)III = imkf = 1 × 0.1 kf = 0.1 kf ⇒ (Tf)III = –0.1kf
⇒ (Tf)II < (Tf)I < (Tf)III
⇒ option [D]

31)
32) The Correct Answer is (B)

33) kf ∝ and kb ∝

34)
= 0.246 atm

35)

PT . YA =

36) Question Explanation:

Relation between P0, PS, N1 and N2

Concept:
This question is based on relative lowering in vapour pressure.

Solution:
R.L.V.P.

P0 – PS = N2P0

Final Answer:
The correct option is (B)

37) Question Explanation:

Calculate degree of dissociation(α) for AgNO3
Concept:
This question is based on Van't Hoff factor(i) for dissociation.
i = 1 + α(n – 1)

Solution:
For AgNO3 → Ag+ +
so no of particles(n) = 2

∵ i = 1 + α(n – 1)
1.835 = 1 + α(2 – 1)
α = .835
% α = 83.5%

Final Answer:
The correct option is (B)

38) Question Explanation:

Calculate mass of ice would have separated.
Concept:
This question is based on depression in freezing point (ΔTf)
Solution:
Initial weight of water = 100 gm
ΔTf = Kf m

ΔTf = Kf m

= .08 kg = 80 gm
Wice = initial mass of H2O – final mass of H2O
= 100 – 80 = 20 gm ice separated
Final Answer:
The correct option is (B)

39)
Molecular weight of Ca(NO3)2 = 164

i = 2.5

Degree of ionisation,
= 0.75 (Ans.)
40) (A) CCl4 + CH3OH

41) πnett = π1 + π2 = i1C1RT + i2C2RT

=
= 0.1 RT + 0.1 RT = 0.2 RT
= 0.2 × 0.08206 × 30
= 4.92 atm

42) i = 1 + 2 × 0.815
i = 2.63

x = 45 gm

43) Explanation:
Given degree of dissociation = 0.7 for

Asked - observed molecular mass to eqm misture

Concept:
This Question is based on Degree of dissociation
Solution:

2NH3(g) N2(g) + 3H2(g) 1 + (2 – 1) (0.7) =

Mobs = 10
Final Answer: Option : (B)

44) A(g) + 2B(s) 2C(g)

t=0 2 4 –
t = teq. 1 2 2

Kp =

45)

46) Molarity of resulting solution ⇒ 0.05 M

on solving V = 100 ml

47) Final moles = 10 m moles + 80 m moles

Final volume = 100 ml + 400 ml + 500 ml = 1 litre
 ∴ Molarity =
∴x=2

48) Conceptual

49) Ans. OMR ANS (8)

ΔTf = Kfm = Kf. = 125

50) KC =

PART-3 : MATHEMATICS

51) RHL = 1 LHL = 0

52) LHL = –2, RHL = 1 ⇒ LHL ≠ RHL

53) (1∞)

54)

⇒
= 1 × π × (1)2 = π

55) = =1

56)
57)

ƒ(x) = x(x – 1)|x(x – 4)(x – 5)|

So ƒ(x) is not differentiable at x = 4 & 5.

58) f(x) is continuous but non-differentiable at infinite points

59) Graph of given function

60) For continuity ƒ(1–) = ƒ(1+)

⇒ a+6=b+2⇒a+4=b
and for differentiability ƒ'(1–) = ƒ'(1+)
⇒ 2a + 6 = 2 ⇒ a = –2

61) Continuous at x = 1

∴ p + q = –2 +
Differentiable at x = 1

62)

63)

at x = 0
ƒ(0–) = ƒ(0+)
 ...(1)
at x = 1
ƒ(1–) = ƒ(1+)
a+1=c–2 ...(2)
at x = 2
ƒ(2–) = ƒ(2+)
4c – 2 = 0 ...(3)
at x = 4
ƒ(4–) = ƒ(4+)

6d = 12 ...(4)

64) Function is differentiable at x = –1 so it also continuous at x = –1

For continuity
p+q=q–p+4⇒p=2
For differentiability
2p(–1) = 2q(–1) + p ⇒ q = 3

65) Clearly ƒ'(–1) = 2

66)

67) y = tan–1

= tan–1 ……(i)

Now, <x<π

or

∴ = – tan
(∵ in II quadrant)

From Eq.(i)

y = tan–1 tan

=π–
 =

68) Let y = f(x) ⇒ x = f–1(y)

then f (x) = x + tan x

⇒ y = f–1(y) + tan
⇒ y = g(y) + tan(g(y)) or x = g(x) + tan (g(x)) ……(i)
Differentiating both sides, then we get
1 = g'(x) + sec2 g(x) · g'(x)

∴ g'(x) = =

69)

70)
Let y = (sinx)x
logy = xlog sin x

+ logsinx

= y(xcotx + logsinx)

= (sinx)x (xcotx + logsinx)

71)
72)

73)
For continuity at '0'

⇒ ⇒ = – ab
⇒ 4 = – ab ⇒ 10 – ab = 14

74) ƒ(x + y) = ƒ(x) + ƒ(y) + x2y + xy2

ƒ'(x) = x2 + 2

75)

⇒ y = sec–1 + sin–1(sin θ)
⇒ y = sec–1 (cosec θ) + θ

⇒ y = sec–1 +θ

y= –θ+θ