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An organ pipe filled with a gas at $27^{\circ} \mathrm{C}$ resonates at $400 \mathrm{~Hz}$ in its fundamental mode. If it is filled with the same gas at $90^{\circ} \mathrm{C}$, the resonance frequency at the same mode will be:
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Verified Answer
The correct answer is:
$440 \mathrm{~Hz}$
As per question, we have;
frequency, $f=\frac{v}{4 l}$
Hence,
$\begin{array}{l}
\Rightarrow \frac{4 l}{f_2} =\frac{v_1}{v_2} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{300}{363}} \\
\Rightarrow [\because \mathrm{V} \propto \sqrt{\mathrm{T}}] \\
\Rightarrow f_2 =\sqrt{\frac{363}{300}} \times 400 \\
\Rightarrow f_2 =440 \mathrm{~Hz}
\end{array}$
frequency, $f=\frac{v}{4 l}$
Hence,
$\begin{array}{l}
\Rightarrow \frac{4 l}{f_2} =\frac{v_1}{v_2} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{300}{363}} \\
\Rightarrow [\because \mathrm{V} \propto \sqrt{\mathrm{T}}] \\
\Rightarrow f_2 =\sqrt{\frac{363}{300}} \times 400 \\
\Rightarrow f_2 =440 \mathrm{~Hz}
\end{array}$
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