The diameter of a wire measured by a screw gauge of least count 0.001… | JEE Main 2026 (2 Apr, Shift 1)

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Errors in Measurement

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young’s modulus is $\alpha \times 10^{9}\text{ N/m}^{2}$ . The value of $\alpha$ is ________.

(Ignore the contribution of the load to Young’s modulus error calculation)

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Answer: (b)

$Y = \dfrac{FL}{(\pi d^2/4)\,l}$ , so $\dfrac{\Delta Y}{Y} = 2\dfrac{\Delta d}{d} + \dfrac{\Delta L}{L} + \dfrac{\Delta l}{l}$ (load error ignored).
$= 2\times\dfrac{0.001}{0.08} + \dfrac{0.1}{150} + \dfrac{0.001}{0.5} = 0.025 + 0.00067 + 0.002 = 0.0277$
$Y = \dfrac{100\times 1.5}{\pi(8\times10^{-4})^2/4 \times 5\times10^{-3}} \approx 5.97\times10^{10}\text{ N/m}^2$
$\Delta Y = 0.0277\times 5.97\times10^{10} \approx 1.65\times10^{9}\text{ N/m}^2$
So $\alpha = 1.65$ , answer (b).

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