A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of 10^5 ~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 ~N / m^2 )

Properties of Matter PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q60 with Solution

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Correct answer: A

For a body under shear,

G = \frac{\text{shear stress}}{\text{shear strain}}

Here,

Shear stress $= \dfrac{F}{A}$
Shear strain $\approx \theta$ for infinitesimally small angular displacement, since $\theta \approx \tan\theta = \frac{x}{L}$

So,

G = \frac{F/A}{\theta}

Hence,

\theta = \frac{F}{AG}

Radius:

r = 4\text{ cm} = 0.04\text{ m}

Area of circular cross-section:

A = \pi r^2 = \pi (0.04)^2 = \pi(0.0016) = 0.0016\pi\, \text{m}^2

Given:

F = 10^5\,\text{N}, \qquad G = 10^{10}\,\text{N/m}^2

Therefore,

\theta = \frac{10^5}{(0.0016\pi)(10^{10})}

\theta = \frac{10^5}{1.6\times 10^7\pi}

\theta = \frac{1}{160\pi}

\theta = \frac{1}{160\pi}

So the correct option is A.