Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 2 April, Morning Shift — Question 1

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 $\alpha \times 10^{9}\mathrm{N} / \mathrm{m}^{2}$ . The value of $\alpha$ is _____

Option A:
1.3
Option B:
1.65
Correct
Option C:
0.13
Option D:
0.25

Answer: B

Step-by-step solution

$\mathrm{Y} = \frac{\mathrm{F}\ell}{\mathrm{A}\Delta\ell}$ $\therefore \frac{\Delta Y}{Y} = \frac{\Delta F}{F} +\frac{\Delta\ell}{\ell} +2\frac{\Delta\ell}{\mathrm{d}} +\frac{\Delta(\Delta\ell)}{\Delta\ell}$ ... $\Delta Y = 1.65\times 10^{9}\mathrm{N / m}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Vernier Calipers and Screw Gauge