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In resonance tube, first and second resonance are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. The third resonance will be obtained at a depth
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The correct answer is:
$117.7 \mathrm{~cm}$
First resonance will occure at $l_1+\mathrm{x}=\frac{\lambda}{4}$
Second resonance will occure at $l_2+\mathrm{x}=\frac{3 \lambda}{4}$
Second resonance will occure at $l_2+\mathrm{x}=\frac{3 \lambda}{4}$
$$
\begin{aligned}
& l_2+\mathrm{x}=3\left(l_1+\mathrm{x}\right) \\
& l_2+\mathrm{x}=3 l_1+3 \mathrm{x} \\
& 2 \mathrm{x}=l_2-3 l_1 \\
& \therefore \quad \mathrm{x}=\frac{l_2-3 l_1}{2} \\
& \quad=\frac{70.2-68.1}{2}=1.05 \mathrm{~cm}
\end{aligned}
$$
$\therefore \quad$ Third resonance occurs at $l_3+\mathrm{x}=\frac{5 \lambda}{4}$
$$
\begin{aligned}
\therefore \quad l_3 & =5\left(l_1+\mathrm{x}\right)-\mathrm{x} \\
& =5 l_1+4 \mathrm{x} \\
& =113.5+4.2 \\
& =117.7 \mathrm{~cm}
\end{aligned}
$$
Second resonance will occure at $l_2+\mathrm{x}=\frac{3 \lambda}{4}$
Second resonance will occure at $l_2+\mathrm{x}=\frac{3 \lambda}{4}$
$$
\begin{aligned}
& l_2+\mathrm{x}=3\left(l_1+\mathrm{x}\right) \\
& l_2+\mathrm{x}=3 l_1+3 \mathrm{x} \\
& 2 \mathrm{x}=l_2-3 l_1 \\
& \therefore \quad \mathrm{x}=\frac{l_2-3 l_1}{2} \\
& \quad=\frac{70.2-68.1}{2}=1.05 \mathrm{~cm}
\end{aligned}
$$
$\therefore \quad$ Third resonance occurs at $l_3+\mathrm{x}=\frac{5 \lambda}{4}$
$$
\begin{aligned}
\therefore \quad l_3 & =5\left(l_1+\mathrm{x}\right)-\mathrm{x} \\
& =5 l_1+4 \mathrm{x} \\
& =113.5+4.2 \\
& =117.7 \mathrm{~cm}
\end{aligned}
$$
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