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Q.
The fourth overtone of an open organ pipe has the same frequency as the second overtone of a close pipe of length $L$. The length of open pipe will be.
Solution:
Fourth overtone of open is its fifth harmonic.
$f=5 \times \frac{v}{2 L_{0}}$
$f'=5 \times \frac{v}{4 L}$
(second over tone of close is its fifth harmonic)
given $f'=f$
$\frac{5 V}{2 L_{0}}=\frac{5 V}{4 L}$
$\therefore 2 L_{0}=4\, L$
$L_{0}=2\, L$