A 100 m long wire having cross-sectional area 6.25 × 10⁻⁴ m² and Young's modulus is 10¹⁰ N m⁻² is subjected to a load of 250 N, then the elongation in the wire will be: (a) 6.25 × 10⁻⁶ m (b) 4 × 10⁻³ m (c) 4 × 10⁻⁴ m (d) 6.25 × 10⁻³ m

Answer: (b). Δ L = FL/AY = 250 × 1006.25 × 10⁻⁴ × 10¹⁰ = 2.5 × 10⁴6.25 × 10⁶ = 4 × 10⁻³ m So the answer is (b).

A 100 m long wire having cross-sectional area 6.25 × 10⁻⁴ m² and… | JEE Main 2023 (24 Jan, Shift 1)

Q11JEE Main 2023 · 24 JAN · SHIFT 1Easy

Young's Modulus & Elongation

A 100 m long wire having cross-sectional area $6.25 \times 10^{-4}\text{ m}^2$ and Young's modulus is $10^{10}\text{ N m}^{-2}$ is subjected to a load of 250 N, then the elongation in the wire will be:

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Answer: (b)

$\Delta L = \frac{FL}{AY}$
$= \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} = \frac{2.5 \times 10^{4}}{6.25 \times 10^{6}}$
$= 4 \times 10^{-3}\ \text{m}$
So the answer is (b).

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