Practice question
Question
A copper wire of length 2.0m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 1×10−4. If the Young's modulus of copper is 1.1×1011N/m2, what is the force applied?
Explanation
Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×1×10−4 = 1.1×107N/m2. Force: F = Stress×A = 1.1×107×2×10−6 = 22N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 22N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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