A) 260 Hz
B) 130 Hz
C) 390 Hz
D) 520 Hz
Correct Answer: A
Solution :
Fundamental frequency of cylindrical open tube \[n=\frac{v}{2L}=390\,\,Hz\] When it is immersed in water it becomes a closed tube of length\[\frac{3}{4}th\]of the initial length. Therefore, its fundamental frequency is \[n=\frac{v}{4\left( \frac{3}{4}L \right)}=\frac{v}{3L}=\frac{2}{3}\left( \frac{v}{2L} \right)\] \[=\frac{2}{3}\,\times 390\,Hz=260\,Hz\]You need to login to perform this action.
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