JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 4)
A 100 m long wire having cross-sectional area $$\mathrm{6.25\times10^{-4}~m^2}$$ and Young's modulus is $$\mathrm{10^{10}~Nm^{-2}}$$ is subjected to a load of 250 N, then the elongation in the wire will be :
$$\mathrm{6.25\times10^{-6}~m}$$
$$\mathrm{4\times10^{-3}~m}$$
$$\mathrm{4\times10^{-4}~m}$$
$$\mathrm{6.25\times10^{-3}~m}$$
Explanation
Elongation in wire $\delta=\frac{\mathrm{F} \ell}{\mathrm{AY}}$
$$ \begin{aligned} & \delta=\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} \\\\ & \delta=4 \times 10^{-3} \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} & \delta=\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} \\\\ & \delta=4 \times 10^{-3} \mathrm{~m} \end{aligned} $$
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