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Physics NEET Questions — Free MCQ Practice

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A source of sound moves away with velocitv of sound from a stationary observer. Then the frequency of the sound heard by the observer :

A.

remains the same

B.

is doubled

C.

is halved

D.

becomes infinity

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Explanation

When a source of sound moves away from a stationary observer, the frequency heard by the observer decreases due to the Doppler effect. The observed frequency is given by f' = f * (v - v_s) / (v + v_s), where f is the source frequency, v is the speed of sound, and v_s is the speed of the source. As the source moves away with the speed of sound, the observed frequency is halved.

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The longitudinal waves can travel :

A.

only in liquids

B.

only in solids

C.

only in gases

D.

In liquids, solids, and gases

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Explanation

4. In liquids, solids, and gases

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$The equation of a wave pulse travelling along x - axis is given as y = \frac{30}{2 + {(x - 20 t)}^{2}} where . x and y are in meters and t is in seconds . The amplitude of the wave pulse is$

A.

5m

B.

20m

C.

15

D.

30m

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Eleven tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tunning forks produces 4 beats per second. The highest frequency is twice that of the lowest. The frequency is twice that of the lowest. The highest and the lowest frequencies (in Hz) are respectively

A.

100 and 50

B.

44 and 22

C.

80 and 40

D.

72 and 30

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Explanation

$Let lowest frequncy = f Then, highest frequncy = f + (11 - 1) \times 4 = f + 40 Now, f + 40 = 2 f \Rightarrow f = 40 Highest frequncy = 80 Hz Lowest frequency = 40 Hz$

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Consecutive frequencies emitted from an organ pipe are 75Hz, 125Hz, 175Hz, the frequency of the 10th overtone will be

A.

275 Hz

B.

175 Hz

C.

525 Hz

D.

575 Hz

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Explanation

$Taking the ratios of the frequencies : 75 : 125 : 175 = (2 \times 1 + 1) \times 25 : (2 \times 2 + 1) \times 25 : (2 \times 3 + 1) \times 25 These are odd harmics . The pipe is closed . Fundamental frequency = 25 Hz Frequency of 10 th overtone = (2 \times 10 + 1) \times 25 = 21 \times 25 = 525 Hz$

TKING

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The sound intensity level at a point 10 m away from a point source is 20dB, then the sound level at a distance 1m from the same source would be

A.

40 dB

B.

30 dB

C.

200 dB

D.

100 dB

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Explanation

$I \propto \frac{1}{a^{2}} \frac{I_{1}}{I_{2}} = {(\frac{a_{2}}{a_{1}})}^{2} = {(\frac{10}{1})}^{2} = 100 S o u n d l e v e l L_{1} = 10 \log_{10} \frac{I_{1}}{I_{o}} S o u n d l e v e l L_{2} = 10 \log_{10} \frac{I_{2}}{I_{o}} ∆ L = L_{1} - L_{2} = 10 \log_{10} (\frac{I_{1}}{I_{2}}) = 10 \log_{10} (100) L - 20 = 10 \log_{10} {(10)}^{2} = 10 \times 2 L = 40 d B$

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In the phenomenon of interference of sound by two coherent sources if difference of intensity at maxima to intensity at minima is 20dB, then the ratio of intensities of the two sources is

A.

$\frac{121}{81}$

B.

$\frac{11}{9}$

C.

$\frac{101}{99}$

D.

$\frac{10}{1}$

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Explanation

Since the difference is in decibel(dB), the difference in sound level (Loudness) is 20 dB

$Loudness L = 10 \log_{10} (\frac{I}{I_{o}}) L_{1} = 10 \log_{10} (\frac{I_{\max}}{I_{o}}) L_{2} = 10 \log_{10} (\frac{I_{\min}}{I_{o}}) ∆ L = L_{1} - L_{2} = 10 \log_{10} (\frac{I_{\max}}{I_{o}}) - 10 \log_{10} (\frac{I_{\min}}{I_{o}}) = 10 \times \log_{10} (\frac{I_{\max}}{I_{\min}}) {logb - loga = \log \frac{b}{a}} ∆ L = 10 \log_{10} (\frac{I_{\max}}{I_{\min}}) \Rightarrow 20 = 10 \log_{10} (\frac{I_{\max}}{I_{\min}}) \Rightarrow \frac{I_{\max}}{I_{\min}} = 100 \frac{I_{\max}}{I_{\min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = 100 \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = 10 \Rightarrow \frac{a_{1}}{a_{2}} = \frac{11}{9} \frac{I_{1}}{I_{2}} = \frac{{a_{1}}^{2}}{{a_{2}}^{2}} = \frac{121}{81}$

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When a sound wave travels from one medium to another, the quantity that remains unchanged is :

A.

speed

B.

amplitude

C.

frequency

D.

wavelength

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Explanation

Frequency is the characteristics property of the source and independent of the medium.

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If a sound source of frequency n approaches an observer with velocity v/4 and the observer approaches the source with velocity v/5, then the apparent frequency heard will be-

A.

(5/8)n

B.

(8/5)n

C.

(7/5)n

D.

(5/7)n

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Explanation

$(2) ν' = ν [\frac{v \pm v_{0}}{v \pm v_{s}}] ν' = ν [\frac{v + \frac{v}{5}}{v - \frac{v}{4}}] = \frac{8 ν}{5}$

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The extension in a string, obeying Hook's law, is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be-

A.

1.22 v

B.

0.61 v

C.

1.50 v

D.

0.75 v

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Explanation

$(1) v = \sqrt{\frac{T}{μ}} \frac{v}{v'} = \sqrt{\frac{kx}{1.5 kx}} \Rightarrow v' = \sqrt{1.5} v \Rightarrow v' = 1.22 v$

Physics MCQs for NEET — Practice Questions with Answers

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