The force required to stretch a wire of cross-section 1 cm² to double its length will be: (Given Young's modulus of the wire = 2 × 10¹¹ N/m²) (a) 1 × 10⁷ N (b) 1.5 × 10⁷ N (c) 2 × 10⁷ N (d) 2.5 × 10⁷ N

Answer: (c). Doubling the length means Δ L = L, so strain = 1. F = Y × A × strain = 2 × 10¹¹ × 10⁻⁴ × 1 F = 2 × 10⁷ N So the answer is (c).

The force required to stretch a wire of cross-section 1 cm² to double… | JEE Main 2022 (28 Jul, Shift 1)

Q18JEE Main 2022 · 28 JUL · SHIFT 1Easy

Young's Modulus & Elongation

The force required to stretch a wire of cross-section $1\text{ cm}^2$ to double its length will be: (Given Young's modulus of the wire $= 2 \times 10^{11}\text{ N/m}^2$ )

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Answer: (c)

Doubling the length means $\Delta L = L$ , so strain $= 1$ .
$F = Y \times A \times \text{strain} = 2 \times 10^{11} \times 10^{-4} \times 1$
$F = 2 \times 10^{7}\ \text{N}$
So the answer is (c).

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