STANDARD Physics Young's Modulus MCQ Question
The area of a cross-section of a steel wire is 0.1 cm² and Young’s modulus of steel is 2 × 10¹¹ N m⁻². The force required to stretch it by 0.1% of its length is
1000 N
2000 N
4000 N
5000 N
Correct Answer
Detailed Explanation
The force required is calculated using the formula F = Y(ΔL/L)A. Given A = 0.1 cm² = 0.1 × 10⁻⁴ m², Y = 2 × 10¹¹ N m⁻², and ΔL/L = 0.1% = 0.1 × 10⁻², the force F = 2 × 10¹¹ × 0.1 × 10⁻² × 0.1 × 10⁻⁴ = 2000 N.
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