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A metallic wire with tension $T$ and at temperature $30^{\circ} \mathrm{C}$ vibrates with its fundamental frequency of $1 \mathrm{kHz}$. The same wire with the same tension but at $10^{\circ} \mathrm{C}$ temperature vibrates with a fundamental frequency of $1.001 \mathrm{kHz}$. The coefficient of linear expansion of the wire is
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Verified Answer
The correct answer is:
$0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
Frequency of wire $n=\frac{1}{2 l} \frac{\sqrt{T}}{m}$ $m=$ mass per unit length
In both the condition same wire is kept same tension.
So,
$$
\begin{aligned}
\frac{n_1}{n_2}=\frac{l_2}{l_1} \Rightarrow \frac{1}{1.001} & =\frac{l_2}{l_1} \\
l_2 & =\frac{l_1}{1.001}
\end{aligned}
$$
Fall in temperature $\Delta t=30-10=20^{\circ} \mathrm{C}$
$$
\begin{aligned}
l_2 & =l_1(1-\alpha \Delta t) \\
\frac{l_1}{1.001} & =l_1(1-\alpha \times 20) \\
1-\alpha \times 20 & =\frac{1}{1.001}=\frac{0.001}{1.001} \\
\alpha=\frac{1}{1001 \times 20} & =0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
In both the condition same wire is kept same tension.
So,
$$
\begin{aligned}
\frac{n_1}{n_2}=\frac{l_2}{l_1} \Rightarrow \frac{1}{1.001} & =\frac{l_2}{l_1} \\
l_2 & =\frac{l_1}{1.001}
\end{aligned}
$$
Fall in temperature $\Delta t=30-10=20^{\circ} \mathrm{C}$
$$
\begin{aligned}
l_2 & =l_1(1-\alpha \Delta t) \\
\frac{l_1}{1.001} & =l_1(1-\alpha \times 20) \\
1-\alpha \times 20 & =\frac{1}{1.001}=\frac{0.001}{1.001} \\
\alpha=\frac{1}{1001 \times 20} & =0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
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