Waves

JEE Main level, numerical answer, medium

95 of 142

Question 95 of 142

A tuning fork is vibrating at 250Hz250Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be _____cmcm. (Take speed of sound in air as 340ms−1340{ms}^{-1} )

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Answer: 34

For closed organ pipe
f=(2n+1)v4lf=(2n+1)\dfrac{v}{4l}
for minimum length, n=0n=0
f0=v4l{f}_{0}=\dfrac{v}{4l}
⇒l=v4f0\Rightarrow l=\dfrac{v}{4{f}_{0}}
=3404×250=34cm=\dfrac{340}{4\times 250}=34cm

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