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A tube of diameter d and of length l is ...

A tube of diameter d and of length l is open a both ends. Its fundamental frequency of resonance is found to be `f_(1)`. One end of the tube is now closed. The lowest frequency of resonance of this closed tube is now `f_(2)`. Taking into consideration the end correction ,`(f_2)/(f_(1)` is

A

`((l +0.6d))/((l+0.3 d))`

B

`((l +0.3d))/(2(l+0.6 d))`

C

`((l +0.6d))/(2(l+0.3 d))`

D

`(1(d +0.3l))/(2(d+0.6l))`

Text Solution

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The correct Answer is:
C
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