Practice question
Question
A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by 0.52mm. If the Young's modulus of steel is 2×1011N/m2, what is the force applied?
Explanation
Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.52×10−3m. F = 2×1011×2×10−6×0.52×10−32.6 = 2082.6 = 80N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 80N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.