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JEE Main 2026 · 5 Apr · Shift 1, Physics question 46

A cube has side length 5 cm and modulus of rigidity 10510^5 N/m². The displacement produced by a force of 10 N in the upper face of cube is ________ mm.

Answer and worked solution

Official answer: 2

Shear stress. 10(0.05)2=4000\dfrac{10}{(0.05)^2}=4000 N/m².

Strain. 4000105=0.04\dfrac{4000}{10^5}=0.04.

Displacement. 0.04×5 cm=0.2 cm=20.04\times5\text{ cm}=0.2\text{ cm}=2 mm.

Answer: 2