As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45 ° with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x × 10⁴ Nm ⁻². The value of x is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 5. Slope (extension)/(load) = tan 45° = 1 m/N Y = (FL)/(AΔ l) = L/A × (1)/(slope) = 0.628π (2 × 10⁻³)² Y = 0.6281.256 × 10⁻⁵ = 5 × 10⁴ N m⁻², so x = 5.

As shown in the figure, in an experiment to determine Young's modulus… | JEE Main 2023 (25 Jan, Shift 1)

Q12JEE Main 2023 · 25 JAN · SHIFT 1Medium

Wire Combinations & Special Cases

As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45 $^\circ$ with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be $x \times 10^{4}$ Nm $^{-2}$ . The value of $x$ is ___________.

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

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Answer: 5

Slope $\frac{\text{extension}}{\text{load}} = \tan 45^\circ = 1\text{ m/N}$
$Y = \frac{FL}{A\Delta l} = \frac{L}{A} \times \frac{1}{\text{slope}} = \frac{0.628}{\pi (2 \times 10^{-3})^2}$
$Y = \frac{0.628}{1.256 \times 10^{-5}} = 5 \times 10^{4}\text{ N m}^{-2}$ , so $x = 5$ .

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