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Physics MCQs for NEET — Practice Questions with Answers

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An observer moves towards a stationary source of sound of frequency n. The apparent frequency heard by him is 2n. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is :

166 m/sec

664 m/sec

332 m/sec

1328 m/sec

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Explanation

$n' = n (\frac{v + v_{O}}{v}) \Rightarrow 2 n = n (\frac{v + v_{0}}{v}) \Rightarrow \frac{v + v_{0}}{v} = 2$

$\Rightarrow v_{O} = v = 332$ m/sec

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A source of sound emitting a note of frequency 200 Hz moves towards an observer with a velocity v equal to the velocity of sound. If the observer also moves away from the source with the same velocity v, the apparent frequency heard by the observer is

50 Hz

100 Hz

150 Hz

200 Hz

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Explanation

Since there is no relative motion between observer and source, therefore there is no apparent change in frequency.

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A source of sound is traveling towards a stationary observer. The frequency of the sound heard by the observer is three times the original frequency. The velocity of sound is v m/sec. The speed of the source will be :

$\frac{2}{3} v$

$\frac{3}{2} v$

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Explanation

$n^{'} = n (\frac{v}{v - v_{S}})$ ⇒ $\frac{n'}{n} = \frac{v}{v - v_{S}}$

$\Rightarrow \frac{v}{v - v_{S}} = 3 \Rightarrow v_{s} = \frac{2 v}{3}$

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A sound source is moving towards a stationary observer with 1/10 of the speed of sound. The ratio of apparent to real frequency is :

10/9

11/10

(11/10)²

(9/10)²

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Explanation

$n' = n (\frac{v}{v - v_{S}}) = n (\frac{v}{v - v / 10})$

⇒ $\frac{n'}{n} = \frac{10}{9}$

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The speed of sound in air at a given temperature is 350 m/s. An engine blows whistle at a frequency of 1200 cps. It is approaching the observer with velocity 50 m/s. The apparent frequency in cps heard by the observer will be :

600

1050

1400

2400

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Explanation

$n'=n (\frac{v}{v - v_{s}}) = 1200 \times (\frac{350}{350 - 50})$ =1400 cps

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The Doppler's effect is applicable for :

Light waves

Sound waves

Space waves

Both (1) and (2)

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Explanation

Doppler’s effect is applicable for both light and sound waves.

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A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

9 : 8

8 : 9

1 : 1

9 : 10

(Speed of sound v = 340 m/s)

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Explanation

When source is approaching the observer, the frequency heard

$n_{a} = (\frac{v}{v - v_{S}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$

When source is receding, the frequency heard $n_{r} = (\frac{v}{v + v_{S}}) \times n$ = $\frac{340}{340 + 20} \times 1000 = 944$

$\Rightarrow n_{a} : n_{r} = 9 : 8$

Short tricks : $\frac{n_{a}}{n_{r}} = \frac{v + v_{S}}{v - v_{S}} = \frac{340 + 20}{340 - 20} = \frac{9}{8} .$

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A source and listener are both moving towards each other with speed v/10, where v is the speed of sound. If the frequency of the note emitted by the source is f, the frequency heard by the listener would be nearly :

1.11 f

1.22 f

1.27 f

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Explanation

When the source and listener are moving towards each other, the apparent frequency heard by the listener is higher due to the Doppler effect. If they are moving with speed v/10 (where v is the speed of sound), the apparent frequency is increased by a factor of (1 + v/10)/(1 - v/10), which is approximately 1.22.

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A motor car blowing a horn of frequency 124vib/sec moves with a velocity 72 km/hr towards a tall wall. The frequency of the reflected sound heard by the driver will be (velocity of sound in air is 330 m/s)

109 vib/sec

132 vib/sec

140 vib/sec

248 vib/sec

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A source of sound of frequency n is moving towards a stationary observer with a speed S. If the speed of sound in air is V and the frequency heard by the observer is n₁, the value of n₁/n is

(V + S)/V

V/(V + S)

(V – S)/V

V/(V – S)

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Explanation

By using $n^{'} = n \frac{v}{v - v_{S}}$ ⇒ $\frac{n_{1}}{n} = (\frac{V}{V - S})$

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