The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm², is 2 × 10¹¹ N/m². The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 1/4 of its value on the earth. The elongation of wire is (Take g on the earth = 10 m/s²): (a) 0.1 cm (b) 1 cm (c) 0.1 mm (d) 1 mm

Answer: (c). Load on the planet: F = mg/4 = 4 × 2.5 = 10 N Elongation Δ l = FL/AY = 10 × 63 × 10⁻⁶ × 2 × 10¹¹ Δ l = 606 × 10⁵ = 10⁻⁴ m = 0.1 mm So the answer is (c) 0.1 mm.

The Young's modulus of a steel wire of length 6 m and cross-sectional… | JEE Main 2023 (1 Feb, Shift 2)

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Young's Modulus & Elongation

The Young's modulus of a steel wire of length $6\text{ m}$ and cross-sectional area $3\text{ mm}^2$ , is $2 \times 10^{11}\text{ N/m}^2$ . The wire is suspended from its support on a given planet. A block of mass $4\text{ kg}$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth. The elongation of wire is (Take $g$ on the earth $= 10\text{ m/s}^2$ ):

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Answer: (c)

Load on the planet: $F = m\frac{g}{4} = 4 \times 2.5 = 10\text{ N}$
Elongation $\Delta l = \frac{FL}{AY} = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}}$
$\Delta l = \frac{60}{6 \times 10^{5}} = 10^{-4}\text{ m} = 0.1\text{ mm}$
So the answer is (c) 0.1 mm.

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