A) 4 m
B) 2 m
C) 8 m
D) 1 m
Correct Answer: A
Solution :
Second overtone of open pipe of length I is \[{{n}_{o}}=\frac{v}{2l}\] ...(i) First overtone of a closed pipe is \[{{n}_{c}}=\frac{v}{4l}=\frac{v}{4\times 2}\] ?..(ii) Equating Eqs. (i) and (ii), we get \[\frac{v}{2l}=\frac{v}{8}\] \[\Rightarrow \] \[l=4\,m\]You need to login to perform this action.
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