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In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
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Verified Answer
The correct answer is:
$1.05 \mathrm{~cm}$
For end correction $x$,
$$
\frac{l_{2}+x}{l_{1}+x}=\frac{3 \lambda / 4}{\lambda / 4}=3
$$
$$
\begin{aligned}
x &=\frac{l_{2}-3 l_{1}}{2} \\
&=\frac{70.2-3 \times 22.7}{2}=1.05 \mathrm{~cm}
\end{aligned}
$$
$$
\frac{l_{2}+x}{l_{1}+x}=\frac{3 \lambda / 4}{\lambda / 4}=3
$$
$$
\begin{aligned}
x &=\frac{l_{2}-3 l_{1}}{2} \\
&=\frac{70.2-3 \times 22.7}{2}=1.05 \mathrm{~cm}
\end{aligned}
$$
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