A closed pipe is in resonance with a tuning f | vedclass

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(B) The frequency of the tuning fork $f$ remains constant. For a closed pipe, the fundamental frequency is given by $f = \frac{v}{4L}$, where $v$ is t

Vedclass · Accepted Answer

$7$

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(B) The frequency of the tuning fork $f$ remains constant. For a closed pipe, the fundamental frequency is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the length of the pipe.Since $f$ is constant, $\frac{v_1}{L_1} = \frac{v_2}{L_2}$, which implies $\frac{L_2}{L_1} = \frac{v_2}{v_1}$.The speed of sound $v$ is proportional to the square root of the absolute temperature $T$ $(v \propto \sqrt{T})$.Thus, $\frac{L_2}{L_1} = \sqrt{\frac{T_2}{T_1}}$.Given $T_1 = 27 + 273 = 300 \,K$ and $T_2 = 7 + 273 = 280 \,K$.$L_1 = 20 \,cm = 200 \,mm$.$L_2 = L_1 \sqrt{\frac{280}{300}} = 200 	imes \sqrt{\frac{28}{30}} = 200 	imes \sqrt{0.9333} \approx 200 	imes 0.966 = 193.2 \,mm$.The change in length $\Delta L = L_1 - L_2 = 200 \,mm - 193.2 \,mm = 6.8 \,mm \approx 7 \,mm$.

$A$ closed pipe is in resonance with a tuning fork at $27^{\circ} C$ when its length is $20 \,cm$. If the pipe is to be in resonance with the same tuning fork at $7^{\circ} C$, then the change in the length of the pipe required is nearly: (in $\,mm$)

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