Question: 40: An organ pipe filled with a gas at 27^{\circ}\mathrm{C} resonates at 400 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 :-
(1) 420 Hz
(2) 440 Hz
(3) 484 Hz
(4) 512 Hz
Answer: Option (2)
Explanation:
For an organ pipe of fixed length vibrating in the same mode, the frequency of resonance is directly proportional to the speed of sound in the gas, i.e. f\propto v.
The speed of sound in a gas is proportional to the square root of its absolute temperature,
so v\propto\sqrt{T}.
Hence, the frequency is also proportional to \sqrt{T}.
Let the initial temperature be T_{1}=27^{\circ}\mathrm{C}=300\,\mathrm{K}
and the final temperature be T_{2}=90^{\circ}\mathrm{C}=363\,\mathrm{K}.
Using the relation \frac{f_{2}}{f_{1}}=\sqrt{\frac{T_{2}}{T_{1}}},
we get \frac{f_{2}}{400}=\sqrt{\frac{363}{300}}.
Thus, f_{2}=400\sqrt{\frac{363}{300}}=400\sqrt{1.21}=400\times1.1=440\,\mathrm{Hz}.
Therefore, the resonance frequency at
90^{\circ}\mathrm{C} is 440\,\mathrm{Hz},
and the correct option is (2).