JEE Main 2020 · Physics · Waves and Sound · medium

At 27^circC the fundamental frequency of a pipe closed at one end is exactly 200 Hz. The pipe is then filled with air at 159^circC. If a tuning fork…

Single correct option

27^\circ\mathrm{C}

the fundamental frequency of a pipe closed at one end is exactly

200\ \mathrm{Hz}

. The pipe is then filled with air at

159^\circ\mathrm{C}

. If a tuning fork of frequency

200\ \mathrm{Hz}

is sounded together with the fundamental of the pipe at this temperature, the number of beats per second is (neglect end correction and expansion of the pipe)

A. $40$
B. $20$
C. $88$
D. $120$

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Answer: (A) $40$

Speed of sound

v \propto \sqrt{T}

with

T

in kelvin:

T_1 = 300\ \mathrm{K}

and

T_2 = 432\ \mathrm{K}

. For a fixed pipe length

f \propto v

, so

f_2 = 200\sqrt{432/300} = 200\sqrt{1.44} = 240\ \mathrm{Hz}

. Beats

= 240 - 200 = 40\ \mathrm{s^{-1}}

. Taking

f \propto T

(not

\sqrt T

) would give

288\ \mathrm{Hz}

and

88

beats.

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