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Waves MCQs for NEET — Physics Questions with Answers

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An open pipe of length l vibrates in the fundamental mode. The pressure variation is maximum at :

1/4 from ends

The middle of the pipe

The ends of the pipe

At 1/8 from ends of pipe middle of the pipe

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Explanation

At the middle of pipe, node is formed.

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The fundamental frequency of pipe is 100 Hz and the other two frequencies are 300 Hz and 500 Hz then :

The pipe is open at both the ends

The pipe is closed at both the ends

One end open and another end is closed

None of the above

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Explanation

For closed organ pipe $n_{1} : n_{2} : n_{3} . . . = 1 : 3 : 5 : . . .$

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The fundamental frequency of an open pipe of length 0.5 m is equal to the frequency of the first overtone of a closed pipe of length l. The value of l_c is (m) :

1.5

0.75

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Explanation

First tone of open pipe = first overtone of closed pipe

⇒ $\frac{v}{2 l_{0}} = \frac{3 v}{4 l_{c}}$ ⇒ $l_{c} = \frac{3 \times 2 \times 0 .5}{4} = 0 .75 m$

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In a closed organ pipe, the frequency of the fundamental note is 50 Hz. The note of which of the following frequencies will not be emitted by it :

50 Hz

100 Hz

150 Hz

None of the above

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Explanation

Only odd harmonics are present.

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On producing the waves of frequency 1000 Hz in a Kundt's tube, the total distance between 6 successive nodes is 85 cm. Speed of sound in the gas filled in the tube is

330 m/s

340 m/s

350 m/s

300 m/s

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Explanation

Distance between six successive node

$= \frac{5 λ}{2} = 85 c m$ ⇒ $λ = \frac{2 \times 85}{5} = 34 c m = 0 .34 m$

Therefore speed of sound in gas

$= n λ = 1000 \times 0 .34 = 340 m / s$

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What is the base frequency if a pipe gives notes of frequencies 425, 255 and 595 and decide whether it is closed at one end or open at both ends :

17, closed

85, closed

17, open

85, open

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Explanation

Let the base frequency be n for closed pipe then notes are $n, 3 n, 5 n . . .$

∴ note $3 n = 255$ ⇒ $n = 85$ , note $5 n = 85 \times 5 = 425$

note $7 n = 7 \times 85 = 595$

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A student determines the velocity of sound with the help of a closed organ pipe. If the observed length for fundamental frequency is 24.7 cm, the length for third harmonic will be :

74.1 cm

72.7 cm

75.4 cm

73.1 cm

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Explanation

$l_{2} = 3 l_{1} = 3 \times 24 .7 = 74 .1 c m$

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In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is :

500

300

330

165

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Explanation

For closed pipe $l_{1} = \frac{v}{4 n}$ ;

$l_{2} = \frac{3 v}{4 n}$ ⇒ $v = 2 n (l_{2} - l_{1})$

⇒ $n = \frac{v}{2 (l_{2} - l_{1})} = \frac{330}{2 \times (0 .49 - 0 .16)} = 500 H z$

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Two closed organ pipes of length 100 cm and 101 cm 16 beats in 20 sec. When each pipe is sounded in its fundamental mode calculate the velocity of sound

303 ms^–1

332 ms^–1

323.2 ms^–1

300 ms^–1

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Explanation

Number of beats per second,

$n = \frac{16}{20} = \frac{4}{5}$ ⇒ $n = n_{1} - n_{2} = \frac{v}{4} (\frac{1}{l_{1}} - \frac{1}{l_{2}})$

⇒ $\frac{4}{5} = \frac{v}{4} (\frac{1}{1} - \frac{1}{1 .01}) = \frac{0 .01 v}{4 \times 1 .01}$

$v = \frac{16 \times 101}{5} = 323 .2 m s^{- 1}$

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An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency 200 Hz. The second harmonic of the same pipes produces 10 beats per second with a source of frequency 420 Hz. The frequency of source is

195 Hz

205 Hz

190 Hz

210 Hz

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Explanation

Initially number of beats per second = 5

∴ Frequency of pipe = 200 ± 5 = 195 Hz or 205 Hz ...(i)

Frequency of second harmonics of the pipe = 2n and number of beats in this case = 10

∴ 2n = 420 ± 10 ⇒ 410 Hz or 430 Hz

⇒ n = 205 Hz or 215 Hz ... (ii)

From equation (i) and (ii) it is clear that n = 205 Hz

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