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Q.

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ____________cm. (Take speed of sound in air as 340 ms-1)

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answer is 34.

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Detailed Solution

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The resonant frequency of a closed organ pipe of length L is f=nv4L
Here, n =odd positive integer v = speed of sound in air
For L to be minimum, n=1∴250=v4L⇒250=3404L⇒L=344×25=0.34m⇒L=34cm
 

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