Practice question
Question
A copper block of dimensions 0.5m×0.3m×0.1m is subjected to a shearing force of 6×104N. If the shear modulus of copper is 4.2×1010N/m2, what is the displacement of the top face?
Explanation
Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.1m. Substitute: Δx = 6×104×0.10.15×4.2×1010 = 60006.3×109≈9.52×10−7m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.52×10−7m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.