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A copper wire of length $1.0 \mathrm{~m}$ and a steel wire of length $0.5 \mathrm{~m}$ having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1 \mathrm{~mm}$. If the Young's modulii of copper and steel are respectively $1.0 \times 10^{11}$ $\mathrm{Nm}^{-2}$ and $2.0 \times 10^{11} \mathrm{Nm}^{-2}$, the total extension of the composite wire is :
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Verified Answer
The correct answer is:
$1.25 \mathrm{~mm}$
$1.25 \mathrm{~mm}$
$\mathrm{Y}_{\mathrm{c}} \times\left(\Delta \mathrm{L}_{\mathrm{c}} / \mathrm{L}_{\mathrm{c}}\right)=\mathrm{Y}_{\mathrm{s}} \times\left(\Delta \mathrm{L}_{\mathrm{s}} / \mathrm{L}_{\mathrm{s}}\right)$
$$
\Rightarrow 1 \times 10^{11} \times\left(\frac{1 \times 10^{-3}}{1}\right)=2 \times 10^{11} \times\left(\frac{\Delta \mathrm{L}_{\mathrm{s}}}{0.5}\right)
$$
$$
\therefore \Delta \mathrm{L}_{\mathrm{s}}=\frac{0.5 \times 10^{-3}}{2}=0.25 \mathrm{~mm}
$$
Therefore, total extension of the composite
$$
\begin{aligned}
& \text { wire }=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{s}} \\
& =1 \mathrm{~mm}+0.25 \mathrm{~m}=1.25 \mathrm{~m}
\end{aligned}
$$
$$
\Rightarrow 1 \times 10^{11} \times\left(\frac{1 \times 10^{-3}}{1}\right)=2 \times 10^{11} \times\left(\frac{\Delta \mathrm{L}_{\mathrm{s}}}{0.5}\right)
$$
$$
\therefore \Delta \mathrm{L}_{\mathrm{s}}=\frac{0.5 \times 10^{-3}}{2}=0.25 \mathrm{~mm}
$$
Therefore, total extension of the composite
$$
\begin{aligned}
& \text { wire }=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{s}} \\
& =1 \mathrm{~mm}+0.25 \mathrm{~m}=1.25 \mathrm{~m}
\end{aligned}
$$
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