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(D) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the mass per unit length.Since t

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$0.5 	imes 10^{-4} /^{\circ} C$

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(D) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the mass per unit length.Since the tension $T$ and the mass per unit length $\mu$ (as the wire is the same) remain constant, we have $n \propto \frac{1}{l}$.Therefore, $\frac{n_1}{n_2} = \frac{l_2}{l_1}$.Given $n_1 = 1 \ kHz$ and $n_2 = 1.001 \ kHz$, we have $\frac{l_2}{l_1} = \frac{1}{1.001}$.Using the thermal expansion formula $l_2 = l_1(1 - \alpha \Delta t)$, where $\Delta t = 30^{\circ} C - 10^{\circ} C = 20^{\circ} C$:$\frac{l_1}{1.001} = l_1(1 - \alpha 	imes 20)$.$1 - 20\alpha = \frac{1}{1.001} \approx 1 - 0.001$.$20\alpha = 0.001$.$\alpha = \frac{0.001}{20} = 0.5 	imes 10^{-4} /^{\circ} C$.

$A$ metallic wire with tension $T$ and at temperature $30^{\circ} C$ vibrates with its fundamental frequency of $1 \ kHz$. The same wire with the same tension but at $10^{\circ} C$ temperature vibrates with a fundamental frequency of $1.001 \ kHz$. The coefficient of linear expansion of the wire is

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