Practice question
Question
A steel wire of length 3.2m and cross-sectional area 5×10−6m2 is stretched by 0.8mm. 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 = 0.8×10−3m. F = 2×1011×5×10−6×0.8×10−33.2 = 8003.2 = 250N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.