The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is __________ mm². ( Given, Y = 2.0 × 10¹¹ Nm⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 40. U = (YAx²)/(2L) ⇒ A = (2UL)/(Yx²) A = 2 × 80 × 202 × 10¹¹ × (0.02)² = (3200)/(8 × 10⁷) = 4 × 10⁻⁵ m² 4 × 10⁻⁵ m² = 40 mm² So the answer is 40 mm².

The elastic potential energy stored in a steel wire of length 20 m… | JEE Main 2023 (13 Apr, Shift 1)

Q5JEE Main 2023 · 13 APR · SHIFT 1Medium

Elastic Potential Energy

The elastic potential energy stored in a steel wire of length $20\text{ m}$ stretched through $2\text{ cm}$ is $80\text{ J}$ . The cross sectional area of the wire is __________ $\text{mm}^2$ .

$($ Given, $Y = 2.0 \times 10^{11}\text{ Nm}^{-2})$

Numerical answer type. Work it out, then open the answer.

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Answer: 40

$U = \dfrac{YAx^2}{2L} \Rightarrow A = \dfrac{2UL}{Yx^2}$
$A = \dfrac{2 \times 80 \times 20}{2 \times 10^{11} \times (0.02)^2} = \dfrac{3200}{8 \times 10^7} = 4 \times 10^{-5}\text{ m}^2$
$4 \times 10^{-5}\text{ m}^2 = 40\text{ mm}^2$
So the answer is 40 mm².

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