A closed organ pipe of length L_1 and an open | vedclass

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(B) For a closed organ pipe,the frequency of the first overtone is given by $f_1 = \frac{3v_1}{4L_1}$.For an open organ pipe,the frequency of the firs

Vedclass · Accepted Answer

$\frac{4 L_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

Vedclass · Answer

(B) For a closed organ pipe,the frequency of the first overtone is given by $f_1 = \frac{3v_1}{4L_1}$.For an open organ pipe,the frequency of the first overtone is given by $f_2 = \frac{2v_2}{2L_2} = \frac{v_2}{L_2}$.Given that the frequencies are equal,we have $\frac{3v_1}{4L_1} = \frac{v_2}{L_2}$.The speed of sound in a gas is $v = \sqrt{\frac{1}{\rho K}}$,where $K$ is the compressibility. Since $K$ is the same for both gases,$v \propto \frac{1}{\sqrt{\rho}}$.Thus,$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$.Substituting this into the frequency equation: $\frac{3}{4L_1} \sqrt{\frac{\rho_2}{\rho_1}} = \frac{1}{L_2}$.Solving for $L_2$,we get $L_2 = \frac{4L_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}$.

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 $(cm)$	$t$. $70$

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