A closed pipe of length $10 \,cm$ has its fundamental frequency half that of the second overtone of an open pipe. The length of the open pipe ......... $cm$
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(c)
$l=0.1 \,m$
Fundamental frequency of pipe $\left(f_1\right)=\frac{v}{4 l}$ or $f_1=\frac{v}{0.4}$
Frequency of $2^{\text {nd }}$ overtone of open pipe $2$ :
$f_2=\frac{3 v}{2 l}$
$2 f_1=f_2$
$2 \times \frac{v}{0.4}=\frac{3 v}{2 l}$
$l=0.3\, m$
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