JEE Main 2026 Apr 6 Shift 1, Physics Q1: Measurements

JEE Main2026Apr 6, Shift 1Physics

The density $ρ$ 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 $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $ρ$ is ________.

A
$0.63%$
B
$0.82%$
C
$0.72%$
D
$0.25%$

Solution

$ρ = \frac{m}{\frac{π}{4} d ^{2} l}$ , so by logarithmic differentiation:

$\frac{Δ ρ}{ρ} = \frac{Δ m}{m} + 2 \frac{Δ d}{d} + \frac{Δ l}{l}$ .

Substitute: $\frac{0.02}{97.42} + 2 \cdot \frac{0.02}{20.20} + \frac{0.05}{8.35} \approx 0.000205 + 0.001980 + 0.005988 = 0.00817$ .

$∴$ percentage error $\approx 0.82%$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q1: Measurements

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