JEE Main 2026 Physics question
A cube has side length 5 cm and modulus of rigidity 10⁵ N/m². The displacement produced by a force of 10 N in the upper face of cube is mm.
Answer: 2
Solution
Shear stress = 10/(0.05)² = 4000 N/m². Shear strain = 4000/10⁵ = 0.04. Displacement = strain × side = 0.04×0.05 = 0.002 m = 2 mm.