Practice question
Question
An aluminium wire of length 2.0m and cross-sectional area 1.5×10−6m2 is stretched by a force of 150N. If the Young's modulus of aluminium is 7×1010N/m2, what is the elongation?
Explanation
Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 150×2.01.5×10−6×7×1010 = 3001.05×105≈2.86×10−3m = 2.86mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.86mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.