Practice question
Question
A copper rod of length 2.0m and cross-sectional area 2.5×10−6m2 is subjected to a tensile force of 500N. If the Young's modulus of copper is 1.1×1011N/m2, what is the strain produced?
Explanation
Stress: Stress = FA = 5002.5×10−6 = 2×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 2×1081.1×1011≈1.82×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82×10−3. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.