Practice question
Question
A brass wire of length 1.9m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 2×10−4. If the Young's modulus of brass is 9×1010N/m2, what is the force applied?
Explanation
Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×2×10−4 = 1.8×107N/m2. Force: F = Stress×A = 1.8×107×2×10−6 = 36N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.