Q119 / 200 · NEET 2022 (Phase 2) · Physics

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:

A
$420 \mathrm{~Hz}$
Correct answer
B
$440 \mathrm{~Hz}$
C
$484 \mathrm{~Hz}$
D
$512 \mathrm{~Hz}$

Solution

As per question, we have; frequency, $f=\frac{v}{4 l}$ Hence, $\begin{array}{l} \Rightarrow \frac{4 l}{f_2} =\frac{v_1}{v_2} \\ \Rightarrow \frac{400}{f_2} =\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\ \Rightarrow \frac{400}{f_2} =\sqrt{\frac{300}{363}} \\ \Rightarrow [\because \mathrm{V} \propto \sqrt{\mathrm{T}}] \\ \Rightarrow f_2 =\sqrt{\frac{363}{300}} \times 400 \\ \Rightarrow f_2 =440 \mathrm{~Hz} \end{array}$

Hence, the correct answer is option: 'B'.

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