JEE Main 2025 Apr 4 Shift 2, Physics Q5: Properties of Solids

JEE Main2025Apr 4, Shift 2Physics

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

A
$\frac{1}{160 π}$
B
$\frac{1}{4 π}$
C
$\frac{1}{40 π}$
D
$\frac{1}{2 π}$

Solution

Shear stress $= \frac{F}{A}$ and the shear modulus relates it to the shear strain $θ$ (the angular displacement of the rod axis): $G = \frac{F / A}{θ}$ .

Cross-sectional area: $A = π r^{2} = π (0.04)^{2} = 16 π \times 1 0^{- 4} m^{2}$ .

$θ = \frac{F}{A G} = \frac{1 0 ^{5}}{16 π \times 1 0 ^{- 4} \times 1 0 ^{10}} = \frac{1 0 ^{5}}{16 π \times 1 0 ^{6}} = \frac{1}{160 π}$ rad.

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JEE Main 2025 Apr 4 Shift 2, Physics Q5: Properties of Solids

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