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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2020 · Paper 1 · Physics · Waves and Sound

A stationary tuning fork is in resonance with an air column in a...

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2ms−12m{s}^{- 1} in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320ms−1,320m{s}^{- 1}, the smallest value of the percentage change required in the length of the pipe is ________.

Correct answer

0

Step-by-step solution

f=v4L1f = \dfrac{v}{4{L}_{1}}

f(320+0320−2)=v4L2f\left(\dfrac{320 + 0}{320 - 2}\right) = \dfrac{v}{4{L}_{2}}

⇒L2L1=318320⇒ \dfrac{{L}_{2}}{{L}_{1}} = \dfrac{318}{320}

⇒L1−L2L1=1−318320=2320⇒ \dfrac{{L}_{1} - {L}_{2}}{{L}_{1}} = 1 - \dfrac{318}{320} = \dfrac{2}{320}

\DeltaLL×100\dfrac{\DeltaL}{L} \times 100 % = \dfrac{20}{32} % = 0 . 62 %

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