Practice question
Question
A steel wire of length 2.0m and cross-sectional area 3×10−6m2 is stretched by 0.4mm. 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 = 0.4×10−3m. F = 2×1011×3×10−6×0.4×10−32 = 2402 = 120N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 120N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.