# CBSE 11TH PHYSICS (Test 1)

Tag: cbse 11th physics
Topic: Waves Tag: CBSE 11TH PHYSICS
Q.1
In wave propagation
A. there is flow of matter but there is movement of disturbance
B. there is flow of matter and there is no movement of disturbance
C. there is no flow of matter but there is movement of disturbance
D. there is no flow of matter and there is no movement of disturbance
Explaination / Solution:

because in wave propagation particles only moves in perodic motion, particles get collide with their neighbouring particles then transfer energy to them and comes back to their normal stage. Now next particle which gain energy get into excited state and starts moving periodically and get collide to next adjacent particle and so on. Thus in wave propagation particles only transfer their kinetic energy and momentum. Hence particles does not move therefore there is no flow of matter but there is movement of disturbance.

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Topic: Waves Tag: CBSE 11TH PHYSICS
Q.2
If y(x, t) = a sin (kx + ωt + φ) represents a wave function
A. travelling in the +ve x direction
B. travelling in the -ve x direction
C. direction of travel is sometimes –ve x and +x at other times
D. the equation does not tell the direction
Explaination / Solution:

On comparing above equation with y(x, t) = a sin (-kx + ωt + φ), we get Distance traveled by wave is along - x axis Hence y(x, t) = a sin (kx + ωt + φ) represents a wave function travelling in the -ve x direction

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Topic: Waves Tag: CBSE 11TH PHYSICS
Q.3
A travelling wave, at a rigid boundary or a closed end,
A. is reflected with a phase difference of 30
B. is reflected with a phase reversal
C. is reflected with a phase difference of 70
D. is reflected with same phase
Explaination / Solution:
No Explaination.

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Topic: Waves Tag: CBSE 11TH PHYSICS
Q.4
Velocity of sound in air is 300 m/s. Then the distance between two successive nodes of a stationary wave of frequency 1000 Hz is.
A. 15 cm
B. 20cm
C. 30cm
D. 10 cm
Explaination / Solution:
No Explaination.

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Topic: Waves Tag: CBSE 11TH PHYSICS
Q.5
A tuning fork of frequency 500 Hz is sounded on a resonance tube. The first & second resonances are obtained at 17 cm and 52 cm. The velocity of sound in m/s is
A. 850
B. 170
C. 350.0
D. 520
Explaination / Solution:
No Explaination.

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Q.6
The pendulum of a wall clock executes
A. an oscillatory motion
B. a translatory motion
C. a rotary motion
D. a circular motion
Explaination / Solution:

It executes simple harmonic motion because in this motion restoring force is directly proportional to the displacement towards the mean position at any time.

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Q.7
The time period T of a spring system of mass m and spring constant k is given by
A. T=πkm
B. T=mk
C.
D. T=2πmk
Explaination / Solution:

If the restoring force of a vibrating or oscillatory system is proportional to the displacement
of the body from its equilibrium position and is directed opposite to the direction of displacement,
the motion of the system is simple harmonic and it is given by

where A, the maximum value of the displacement, is called the amplitude of the motion. If
T is the time for one complete oscillation, then
x(t + T) = x(t)

or
As angular frequency   is given by  and

Then,time period of oscillation of mass m is given by

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Q.8
What is constant in simple harmonic motion?
A. Potential energy
B. Time period
C. Restoring force
D. Kinetic motion
Explaination / Solution:

The time period of the SHM is given by  where 'm' be the mass of the body (constant) , 'k' restoring force constant as T depends on 'm' and 'k' and they are constant for the system, so the corresponding Time period of the motion is Constant.

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Q.9
A spring has a certain mass suspended from it and its period for vertical oscillations is  . The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now  .The ratio of  is
A. 2
C.

D.
2
Explaination / Solution:
No Explaination.

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Q.10
Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?
A. 30
B. 60
C. 45
D. 90
Explaination / Solution:

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period

As the two pendulums start at the same time in simple harmonic motion from the mean position,their initial phase angles are zero

The bigger pendulum completes its one oscillation in time  and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle 2π

If ω represents the angular velocity of the reference point associated with SHM of first pendulum of time period T,then the phase of the this pendulum after  s willbe

Hence the phase difference

= 90

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