Class 11 Sample Paper 2022 Solution Physics Term 2

Directorate of Education, GNCT of Delhi
Suggestive answer key
Practice Paper – I
Class – XI
(Code: 042)

खंड - अ
SECTION – A
प्र. स.
Q. No.
Q1.
Iron is more elastic because the more the young's modulus more the
elasticity.
Since
Young’s modulus = Stress/Strain
i.e. Deformation(strain) is less in iron as compared to rubber hence young
modulus is greater in iron thus more elastic.
Q2. Correct statement of Pascal’s law
Names of any two applications
OR
Correct statement of Stoke’s law
Names of any two applications of Stoke’s law

Q3.
Correct definition of latent heat

खंड – ब
SECTION – B
Q4. a. Adiabatic process
b. Isothermal process
c. First law of thermodynamics

Q5. a) At zero kelvin
b) Rectangular hyperbola

c) We know that P=1/3 ρ v2

= 1 /3( M/ V) v2

But v2 ∝ T

∴ P ∝ MT/V

As both T and V remain unchanged but mass M is doubled, so the pressure
of mixture gets doubled, i.e., it is equal to 2P.

So pressure of mixed gases get doubled.

Q6.
a) Derivation of formula with figure and steps.
b) The time period of a simple pendulum is
T=2π √(L /g)
where L is the length of the pendulum.
or L= gT2 /4π 2
The time period of the simple pendulum which ticks seconds is 2s.
∴T=2s
Substituting in (i), we get

L= (9.8)(2)2/ 4×(3.14)2 = 0.992m ≃1m

Q7.
a) Velocity of sound is directly proportional to square root of the temperature
of the air.
As temperature of warm air is more than temperature of cold air, velocity of
sound is more in warm air than cold air.
i) Frequency of the ultrasonic sound, f=100 kHz=105Hz
Speed of sound in air, Va=340m/s
The wavelength (λr) of the reflected sound is given by
λr =Va / f
=340/105
=3.4×10-3 m.

ii) Frequency of the ultrasonic sound, f= 100 kHz =105Hz
Speed of sound in water, Vw =1486 m/s
λt =Vw/ f

The wavelength of the transmitted sound is given by
λt=1486/105
=1.49×10−2m.

Q8 Correct statement
Labelled diagram with proof
Limitations

OR
As velocity increases viscous force (fv=6πηRv) also increases and a point
comes when viscous force become equal to gravitational force (mg) . At this

point there is no net force acting on the body due to opposite direction of the
two forces. Hence velocity of the object stops increasing. This highest velocity
which object has attained is called terminal velocity.

Derivation of an expression for terminal velocity.

Q9 Correct definition of triple point of water.
The triple point of water is a unique temperature and does not change with
pressure and other external factors. The melting point and boiling point of
water vary with pressure. That is, there is no fixed melting and boiling point
of water.

Q10 The frequency of external periodic force is different from the natural
frequency of the oscillator in case of forced oscillation but in resonance two
frequencies are equal.

2) Since x = A cos (ω t +θ)

Velcoity, v= dx/dt = −A ω sin(ωt+θ)
At t=0, x = x0
x0 = A cos θ ...(i)
and, dx/ dt = −v0 = A ω sin θ
A sin θ = v0/ω ...(ii)

Squaring and adding equations (i) and (ii), we get:
A2(cos2θ+sin2θ)=x02+(v02 / ω2)
∴ A=[ x02+( v02 / ω2)2]1/2

Q11. 1) Only (ii) represents S.H.M . For SHM, acceleration should be directly
proportional to displacement and sign should be negative. Hence, (ii)
represents SHM.

2) The amplitude is A=1/2 m= 0.5m.

maximum velocity is given by

Vmax = ωA =200×0.5=100 m/min

OR

Let original freq. of sitar string A be na and original freq. of sitar string B
be nb
As number of beats / sec. =6

∴nb=na ±6

=324±6

=330or318Hz
When tension in A is reduced, its frequency reduces (∵n∝√T)

As number of beats /sec decreases to 3 therefore, frequency

of B=324−6=318Hz

खंड – स
SECTION – C
Q12 a) ( i )
b) ( ii )
c) ( i )
d) ( ii )
e) ( i )