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An organ pipe filled with a gas at $27^\circ C$ resonates at 400 Hz in its fundamental mode. If it is filled with the same gas at $90^\circ C$, the resonance frequency at the same mode will be:
Detailed Solution
$f=\frac{v}{\lambda}$ with $\lambda$ fixed, and $v\propto\sqrt{T}$. $\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{300}{363}}=\frac{1}{1.1} \Rightarrow f_2=400\times1.1=440$ Hz.