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Units and Measurements Part 3

Section: Physics

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Question : 16 of 30

Marks: +1, -0

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 α is _________.

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

[JEE Main 2 apr 2026 Shift 1]

1.3
1.65
0.13
0.25

Solution:

Y = \; \frac{F \ell}{A \Delta \ell}

\therefore \; \frac{\Delta Y}{Y} = \; \frac{\Delta F}{F} + \; \frac{\Delta \ell}{\ell} + 2 \; \frac{\Delta d}{d} + \; \frac{\Delta(\Delta \ell)}{\Delta \ell}

\left[ A = \pi \; \frac{d^{2}}{4} \Rightarrow \; \frac{\Delta A}{A} = \; \frac{2 \Delta d}{d} \right]

Since F is known exactly

\Rightarrow \Delta F = 0

\; \frac{\Delta Y}{Y} = \; \frac{0.1}{150} + \; \frac{2 \times 0.001}{0.08} + \; \frac{0.001}{0.5} = 0.0277

\therefore \Delta Y = 0.0277 Y

Y = \frac{100 \times 1.5}{ \frac{\pi}{4} \left( \frac{0.08}{100} \right)^{2} \times 0.5 \times 10^{-2} } = 5.97 \times 10^{10} \, \text{N/m}^{2}

\therefore \Delta Y = 0.0277 \times 5.97 \times 10^{10} \, \text{N/m}^{2} = 1.65 \times 10^{9} \, \text{N/m}^{2}

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