Practice question
Question
An aluminium wire of length 1.7m and cross-sectional area 2×10−6m2 is stretched by 0.34mm. If the Young's modulus of aluminium is 7×1010N/m2, what is the force applied?
Explanation
Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.34×10−3m. F = 7×1010×2×10−6×0.34×10−31.7 = 47.61.7≈28N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.