n_{1} is the frequency of a pipe closed at on | vedclass

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(A) The fundamental frequency of a pipe closed at one end is $n_{1} = \frac{v}{4 l_{1}}$,which gives $l_{1} = \frac{v}{4 n_{1}}$.The fundamental frequ

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$\frac{n_{1} n_{2}}{n_{2}+2 n_{1}}$

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(A) The fundamental frequency of a pipe closed at one end is $n_{1} = \frac{v}{4 l_{1}}$,which gives $l_{1} = \frac{v}{4 n_{1}}$.The fundamental frequency of a pipe open at both ends is $n_{2} = \frac{v}{2 l_{2}}$,which gives $l_{2} = \frac{v}{2 n_{2}}$.When both pipes are joined end to end,the new pipe is closed at one end and open at the other,with a total length $L = l_{1} + l_{2}$.The fundamental frequency $n$ of this new closed pipe is given by $n = \frac{v}{4 L} = \frac{v}{4 (l_{1} + l_{2})}$.Substituting the values of $l_{1}$ and $l_{2}$: $\frac{1}{4 n} = \frac{1}{4 n_{1}} + \frac{1}{2 n_{2}}$.Multiplying by $4$: $\frac{1}{n} = \frac{1}{n_{1}} + \frac{2}{n_{2}} = \frac{n_{2} + 2 n_{1}}{n_{1} n_{2}}$.Therefore,$n = \frac{n_{1} n_{2}}{n_{2} + 2 n_{1}}$.

List-$I$	List-$II$
$A$. Wavelength of sound $(cm)$	$p$. $1$
$B$. Height of liquid column at $II$ resonance $(cm)$	$q$. $2$
$C$. Speed of sound $(m/s)$	$r$. $340$
$D$. End correction $(cm)$	$s$. $68$
$E$. Minimum level of liquid column at resonance $(cm)$	$t$. $70$

$n_{1}$ is the frequency of a pipe closed at one end and $n_{2}$ is the frequency of a pipe open at both ends. If both are joined end to end,find the fundamental frequency of the closed pipe so formed.

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