The elongation of copper wire of cross-sectional area $3.5 \mathrm{~mm}^2$, in the figure shown, is $\left(\mathrm{Y}_{\text {copper }}=10 \times 10^{10} \mathrm{Nm}^{-2} \text { and } g=10 \mathrm{~ms}^{-2}\right)$

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Aicharya · Accepted Answer

Tension in copper wire $\mathrm{T}_2=70 \mathrm{~N}$ $\therefore$ Elongation in copper wire, $\begin{aligned} & \mathrm{Sl}=\frac{\mathrm{T}_2 \mathrm{I}_2}{\mathrm{Y}_2 \mathrm{~A}_2} \\ & =\frac{70 \times 0.5}{10 \times 10^{10} \times 3.5 \times 10^{-6}} \\ & =10^{-4} \mathrm{~m} \end{aligned}$

The elongation of copper wire of cross-sectional area $3.5 \mathrm{~mm}^2$, in the figure shown, is…

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