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A rod is found to be $0.05 \mathrm{~cm}$ longer at $40^{\circ} \mathrm{C}$ than it is at $10^{\circ} \mathrm{C}$. The length of the rod at $0^{\circ} \mathrm{C}$ is
(coefficient of linear expansion of the material of the rod $\left.=1.5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$
Options:
(coefficient of linear expansion of the material of the rod $\left.=1.5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$
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1693 Upvotes
Verified Answer
The correct answer is:
$111.1 \mathrm{~cm}$
$\begin{aligned}
& \mathrm{L} \text { at } 10^{\circ} \mathrm{C}=\mathrm{L}_0(1+10 \alpha) \\
& \mathrm{L} \text { at } 40^{\circ} \mathrm{C}=\mathrm{L}_0(1+40 \alpha) \\
& \mathrm{L}_{40^{\circ} \mathrm{C}}=0.05+\mathrm{L}_{10^{\circ} \mathrm{C}} \\
& \mathrm{L}_0+\mathrm{L}_0 \alpha \times 40=0.05+\mathrm{L}_0+10 \mathrm{~L}_0 \alpha \\
& 30 \mathrm{I}_0 \alpha=0.05 \\
& \mathrm{~L}_0=\frac{0.05}{30 \alpha}
\end{aligned}$
The length of the rod at $0^{\circ} \mathrm{C}$
$=\frac{0.05}{30 \times 1.5 \times 10^{-5}}=111.1 \mathrm{~cm}$
& \mathrm{L} \text { at } 10^{\circ} \mathrm{C}=\mathrm{L}_0(1+10 \alpha) \\
& \mathrm{L} \text { at } 40^{\circ} \mathrm{C}=\mathrm{L}_0(1+40 \alpha) \\
& \mathrm{L}_{40^{\circ} \mathrm{C}}=0.05+\mathrm{L}_{10^{\circ} \mathrm{C}} \\
& \mathrm{L}_0+\mathrm{L}_0 \alpha \times 40=0.05+\mathrm{L}_0+10 \mathrm{~L}_0 \alpha \\
& 30 \mathrm{I}_0 \alpha=0.05 \\
& \mathrm{~L}_0=\frac{0.05}{30 \alpha}
\end{aligned}$
The length of the rod at $0^{\circ} \mathrm{C}$
$=\frac{0.05}{30 \times 1.5 \times 10^{-5}}=111.1 \mathrm{~cm}$
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