Class 11 Physics Mcqs

Q:

A man is moving towards a wall with a source of sound in his hand. Frequency of source = 50Hz. Speed of man = 8m/s. If the man sends one pulse of sound, what will be the frequency heard by him after sound reflects off the wall? Speed of sound = 330m/s.

A) 53.86Hz B) 52.48Hz
C) 51.24Hz D) 52.45Hz
 
Answer & Explanation Answer: B) 52.48Hz

Explanation: First we find frequency heard by wall.
That will be 50(330)/(330-8) = 51.24Hz.
Now, this will act as source and man will be the listener.
So, the frequency heard by the man
= 51.24(330+8)/330
= 52.48Hz.

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98
Q:

If both the source of sound and observer are moving, the apparent frequency will be different from the actual frequency. True or False?

A) True B) False
 
Answer & Explanation Answer: B) False

Explanation: If relative motion between source and observer is zero there will be no change in frequency heard by the observer. Apparent frequency only varies in case of relative motion. So the given statement is false.

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55
Q:

A man jumps from a building with a source of sound. When he reaches halfway down, he starts the siren of frequency 100Hz. A person standing below him on the ground hears the first sound to be of frequency 120Hz. What is the height of the building? Speed of sound = 330m/s. Take g=9.81m/s².

A) 154m B) 308m
C) 77m D) 500m
 
Answer & Explanation Answer: B) 308m

Explanation: We can find the speed corresponding to the apparent frequency and then use that to find the height of the building.
120 = 100(330)/(330-v) v = 330 – 3300/12 = 55m/s.
Using, v² = 2gh, we can find the height covered by the man. The double of this value will be the height of the building as he is mid-way.
55² = 2*9.81*h h = 154m Height of building = 2h = 308m.

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569
Q:

A source of frequency 50Hz is moving towards an observer with a deceleration of 2m/s². If its initial speed is 20m/s, what will be the maximum frequency heard by the observer? Initial distance between source and observer is 100m. Speed of sound = 330m/s.

A) 47.14Hz B) 0
C) 53.22Hz D) 53.03Hz
 
Answer & Explanation Answer: C) 53.22Hz

Explanation: Apparent frequency only depends on velocity and not the distance between source and observer. So, maximum frequency will be heard when speed is maximum, which is at the start. F = 50(330)/(330-20)
= 53.22Hz.

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126
Q:

A source moves towards a stationary observer with a speed of 10m/s. If the frequency of the source is 10Hz what will be the wavelength of sound heard by the observer? Speed of sound = 330m/s.

A) 32m B) 34m
C) 33m D) 31m
 
Answer & Explanation Answer: A) 32m

Explanation: Apparent frequency F = f(330+v?)/(330+v?)
= 10(330)/(330-10) = 10.31Hz.
Wavelength = 330/F
= 330/10.31
= 32.007m = 32m.

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70
Q:

A source of sound moves towards an observer. What happens to the speed of sound in the medium?

A) Increases B) Decreases
C) Remains the same D) Depends on speed with which source moves
 
Answer & Explanation Answer: C) Remains the same

Explanation: Speed of sound in the medium depends on the properties of the medium like bulk modulus and density, not on the speed of source creating the sound.

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48
Q:

A train with a frequency ‘f’. What will be the frequency heard by a person sitting in the train? Speed of the train is ‘v’.

A) f B) f(330+v)/330
C) fv/(330-v) D) fv/(330+v)
 
Answer & Explanation Answer: A) f

Explanation: There is no relative motion between the source and observer, thus the frequency heard by the person will be the original frequency of the whistle.

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50
Q:

An observer is standing stationary in air moving at a speed of 2m/s. A source of frequency 20Hz is moving in the direction of wind with speed 10m/s towards the observer. What will be the frequency as heard by the observer? Speed of sound in air is 330m/s.

A) 20.62Hz B) 19.40Hz
C) 21.68Hz D) 22.63Hz
 
Answer & Explanation Answer: A) 20.62Hz

Explanation: As the air is moving, we can solve this question from the air’s frame of reference. From the air’s frame of reference, the observer will be going towards the observer at 2m/s and the source will be coming towards the observer at 10-2 = 8m/s.
F = f(v+v?)/(v+v?)
= 20(330+2)/(330-8)
= 20*332/322 = 20.62Hz.

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