Sound(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

SOUND MCQs

Total Questions : 31 | Page 1 of 4 pages

A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is 320 m/s) will be equal to

Discuss Question

Answer: Option A. -> 6
:
A
Apparent frequency due to train which is coming in is

m_{1} = \frac{v}{v - v_{s}} n

Apparent frequency due to train which is leaving is

m_{2} = \frac{v}{v - v_{s}} n

So the number of beats is

n_{1} - n_{2} = (\frac{1}{316} - \frac{1}{324}) 320 \times 240 \Rightarrow n_{1} - n_{2} = 6

Question 2.

The intensity of a sound wave gets reduced by 20% on passing through a slab. The reduction in intensity on passage through two such consecutive slabs is

Discuss Question

Answer: Option B. -> 36%
:
B
Intensity after passing through one slab

I^{'} = [I - \frac{20}{100} \times I] = [I - \frac{1}{5}] = \frac{41}{5}

So, intensity after passing through two slabs

I "= [I - \frac{20}{100} \times I^{'}] = \frac{4 I^{'}}{5} = \frac{161}{25} ∴ % d e c r e a s e = [\frac{(I = - \frac{161}{5})}{I}] \times 100 = 36 %

Question 3.

Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a constant velocity u. If the speed of sound is 340 m/s, what must be the value of u so that he hears
10 beats per second?

2.0 m/s
2.5 m/s
30 m/s
3.5 m/s

Discuss Question

Answer: Option B. -> 2.5 m/s
:
B
Apparent frequency due to source A is

n^{'} = \frac{v - u}{v} \times n

Apparent frequency due to source B is

n "= \frac{v + u}{v} \times n

∴ n " - n^{'} = \frac{2 u}{v} \times n = 10

∴ u = \frac{10 v}{2 n} = \frac{10 \times 340}{2 \times 680} = 2.5 m / s

Question 4.

Two factories are sounding their sirens at 800 Hz. A man goes from one factory to the other at a speed of 2 m/s. The velocity of sound is 320 m/s. The number of beats heard by the person in 1 s will be

Discuss Question

Answer: Option D. -> 10
:
D
When the man is approaching the factory,

n^{'} = (\frac{v + v_{0}}{v}) n = (\frac{320 + 2}{320}) 800 = (\frac{322}{320}) 800

When the man is going away from the factory,

n "= (\frac{v - v_{0}}{v}) n = (\frac{320 - 2}{320}) 800 = (\frac{318}{320}) 800 ∴ n^{'} - n "= (\frac{320 - 318}{320}) 800 = 10 H z

Question 5.

One train is approaching an observer at rest and another train is receding from him with the same velocity 4 m/s. Both the trains blow whistles of same frequency of 243 Hz. The beat frequency in Hz as heard by the observer is (speed of sound in air is 320 m/s)

Discuss Question

Answer: Option B. -> 6
:
B
When the train is approaching,

n_{1} = \frac{v}{v - v_{s}} \times n = \frac{320}{320 - 4} \times 243 = \frac{80}{79} \times 243

When the train is receding,

n_{2} = \frac{v}{v + v_{s}} \times n = \frac{320}{324} \times 243 = \frac{80}{81} \times 243

Beat frequency is

n = n_{1} - n_{2} = 80 \times 243 (\frac{1}{79} - \frac{1}{81}) = 6 H z

Question 6.

A train has just completed a U-curve in a track which is a semi-circle. The engine is at the forward end of the semi-circular part of the track while the last carriage is at the rear end of the semi-circular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is 340 m/s. Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30 m/s, is

219 Hz
188 Hz
200 Hz
181 Hz

Discuss Question

Answer: Option C. -> 200 Hz
:
C
The Doppler formula holds for non-collinear motion if v_s and v₀ are taken to be the resolved component along the line of slight, In this case, we have

A Train Has Just Completed A U-curve In A Track Which Is A S...

v_{0} = - v_{t} s i n 45^{\circ} = - \frac{30}{\sqrt{2}} m / s

v_{s} = - v_{t} s i n 45^{\circ} = \frac{30}{\sqrt{2}} m / s

We have, v = 340 m/s, n = 200 Hz. The apparent frequency n’ is given by

n^{'} = n [\frac{v - v_{0}}{v - v_{s}}] = 200 [\frac{340 + (\frac{30}{\sqrt{2}})}{340 + (\frac{30}{\sqrt{2}})}] = 200 H z

Question 7.

Two sound sources are moving in opposite directions with velocities $v_{1}$ and $v_{2} (v_{1} > v_{2})$ . Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of $v_{1} - v_{2}$ so that the beat frequency observed by the observer is 6 Hz? Speed of sound v = 300 m/s. Given that $v_{1}$ and $v_{2}$ ≪ v.

1 m/s
2 m/s
3 m/s
4 m/s

Discuss Question

Answer: Option B. -> 2 m/s
:
B

f_{1} = 900 (\frac{300}{300 + v_{1}}) ≅ 900 {(1 + \frac{r_{1}}{300})}^{- 1} 900 - 3 v_{1}

Similarly,

f_{2} = 900 (\frac{300}{300 + v_{2}}) = 900 - 3 v_{2} f_{2} - f_{1} = 6 ∴ 3 (v_{1} - v_{2}) = 6 ∴ 3 (v_{1} - v_{2}) = 6 o r v_{1} - v_{2} = 2 m / s

Question 8.

An engine running at speed v/10 sounds a whistle of frequency 600 Hz. A passenger in a train coming from the opposite side at speed v/15 experiences this whistle to be of frequency f. If v is speed of sound in air and there is no wind, f is nearest to

711 Hz
630 Hz
580 Hz
510 Hz

Discuss Question

Answer: Option A. -> 711 Hz
:
A
As the source and the observer are approaching one another, so n’ would be larger.

f = (\frac{v + \frac{v}{15}}{v - \frac{v}{10}}) 600 = 711 H z

Question 9.

A source of sound produces waves of wavelength 60 cm when it is stationary. If the speed of sound in air is 320 m/s and source moves with speed 20 m/s, the wavelength of sound in the forward direction will be approximately:

56.2 cm
60.8 cm
64.4 cm
68.6 cm

Discuss Question

Answer: Option A. -> 56.2 cm
:
A
Apparent frequency is given by:

n^{'} = n (\frac{v}{v - v_{s}})

For a medium, frequency is inversely proportional to wavelength.

λ^{'} = (\frac{v - v_{s}}{v}) λ

λ^{'} = (\frac{320 - 20}{320}) 60

λ^{'} = 56.25 c m

Question 10.

The apparent frequency of the whistle of an engine changes in the ratio of 6:5 as the engine passes a stationary observer. If the velocity of sound is 330 m/s, then the velocity of the engine is