Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 6 April, Morning Shift — Question 1

The density $\rho$ of a uniform cylinder is determined by measuring its mass m, length l and diameter d. The measured values of m, l and d are $97.42 \pm 0.02 \mathrm{g}, 8.35 \pm 0.05 \mathrm{mm}$ and $20.20 \pm 0.02 \mathrm{mm}$ respectively. Calculated percentage fractional error in $\rho$ is

Option A:
63 %
Option B:
82 %
Correct
Option C:
72 %
Option D:
25 %

Answer: B

Step-by-step solution

$\rho = m/(\frac{\pi}{4}d^2\ell)$ . $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{2\Delta d}{d} + \frac{\Delta\ell}{\ell} = \frac{0.02}{97.42} + 2\times\frac{0.02}{20.20} + \frac{0.05}{8.35} = 0.0082 = 0.82\%$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis