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CGP EDU Academic Team
Published on: September 12, 2026
A steel wire of lm long and $1 \mathrm{mm}^2$ cross section area is hang from rigid end. When weight of 1kg is hung from it then change in length will be (given $Y = 2 \times 10^{11} \mathrm{N/m}^2$
Text Solution
Verified by ExpertsThe correct answer is:
C
$l = \frac{\mathrm{MgL}}{\mathrm{YA}} = \frac{1 \times 10 \times 1}{2 \times 10^{11} \times 10^{-6}} = 0.05 \text{ mm}$
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