JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Elasticity & SHM

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of

1 0^{5} N

at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement

θ

of the rod axis from its original position would be : (shear moduli,

G = 1 0^{10} N / m^{2}

)

Answer: (A)

Solution

The shear stress on the top face is

\frac{F}{A}

with

A = π r^{2} = π (0.04)^{2} = 16 π \times 1 0^{- 4} m^{2}

, giving

\frac{F}{A} = \frac{1 0 ^{5}}{16 π \times 1 0 ^{- 4}} = \frac{1 0 ^{9}}{16 π} N/m^{2}

. The shear modulus relates this stress to the angular displacement through

G = \frac{F / A}{θ}

, so

θ = \frac{F / A}{G} = \frac{1 0 ^{9}}{16 π \times 1 0 ^{10}} = \frac{1}{160 π} rad

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option