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In a pipe opened at both ends n(1) and n...

In a pipe opened at both ends `n_(1)` and `n_(2)` be the frequencies corresponding to vibrating lengths `L_(1)` and `L_(2)` respectively .The end correction is

A

`(n_1l_1-n_1l_2)/(2(n_1-n_2))`

B

`(n_2l_2-n_1l_1)/(2(n_2-n_1))`

C

`(n_2l_2-n_1l_1)/(2(n_1-n_2))`

D

`(n_1l_1-n_2l_2)/(n_1-n_2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`v=2n_1(l_1+2e)` For first open pipe
`v=2n_2(l_2+2e)` For second open pipe
`2n_1(l_1+2e) = (2n_2 (l_2 +2e)`
`n_1l_1+2n_1e = n_2l_2 +2n_2e`
`n_2l_2 - n_1l_1 = 2e(n_1 - n_2)`
`e=1/2(n_2l_2-n_1l_1)/(n_1-n_2)`
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