JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 10)
A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $10^5 \mathrm{~N}$ at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $\theta$ of the rod axis from its original position would be : (shear moduli, $G=10^{10} \mathrm{~N} / \mathrm{m}^2$ )
$1 / 160 \pi$
$1 / 2 \pi$
$1 / 4 \pi$
$1 / 40 \pi$
Explanation
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$$\begin{aligned} & \text { Shear moduli }=\frac{\sigma_{\text {shear }}}{\theta} \\ & 10^{10}=\frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta} \\ & \theta=\frac{1}{160 \pi} \text { Radian } \end{aligned}$$
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