The maximum percentage error in the measurement of density of a wire is [Given, mass of wire = (0.60 ± 0.003) g radius of wire = (0.50 ± 0.01) cm length of wire = (10.00 ± 0.05) cm] (a) 7 (b) 8 (c) 5 (d) 4

Answer: (c). ρ = (m)/(π r² l), so (Δρ)/(ρ) = (Δ m)/(m) + 2(Δ r)/(r) + (Δ l)/(l) = (0.003)/(0.60) + 2·(0.01)/(0.50) + (0.05)/(10.00) = 0.5% + 4% + 0.5% So the answer is (c) 5.

The maximum percentage error in the measurement of density of a wire… | JEE Main 2025 (22 Jan, Shift 2)

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Errors in Measurement

The maximum percentage error in the measurement of density of a wire is

[Given, mass of wire $= (0.60 \pm 0.003)\text{ g}$

radius of wire $= (0.50 \pm 0.01)\text{ cm}$

length of wire $= (10.00 \pm 0.05)\text{ cm}]$

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Answer: (c)

$\rho = \frac{m}{\pi r^2 l}$ , so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$
$= \frac{0.003}{0.60} + 2\cdot\frac{0.01}{0.50} + \frac{0.05}{10.00} = 0.5\% + 4\% + 0.5\%$
So the answer is (c) 5.

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