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 ± 0.02 g, 8.35 ± 0.05 mm and 20.20 ± 0.02 mm, respectively. Calculated percentage fractional error in ρ is ______. (a) 0.63% (b) 0.82% (c) 0.72% (d) 0.25%

Answer: (b). ρ = (m)/(π (d/2)² l) = (4m)/(π d² l) ⇒ (Δ ρ)/(ρ) = (Δ m)/(m) + 2(Δ d)/(d) + (Δ l)/(l) (Δ m)/(m) = (0.02)/(97.42) = 0.021%, 2(Δ d)/(d) = (0.04)/(20.20) = 0.198%, (Δ l)/(l) = (0.05)/(8.35) = 0.599% Total = 0.021 + 0.198 + 0.599 ≈ 0.82% So the answer is (b) 0.82%.

The density ρ of a uniform cylinder is determined by measuring its… | JEE Main 2026 (6 Apr, Shift 1)

Q30JEE Main 2026 · 6 APR · SHIFT 1Medium

Errors in Measurement

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 $97.42 \pm 0.02\text{ g},\ 8.35 \pm 0.05\text{ mm}$ and $20.20 \pm 0.02\text{ mm}$ , respectively. Calculated percentage fractional error in $\rho$ is ______.

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Answer: (b)

$\rho = \frac{m}{\pi (d/2)^2 l} = \frac{4m}{\pi d^2 l} \Rightarrow \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$
$\frac{\Delta m}{m} = \frac{0.02}{97.42} = 0.021\%$ , $2\frac{\Delta d}{d} = \frac{0.04}{20.20} = 0.198\%$ , $\frac{\Delta l}{l} = \frac{0.05}{8.35} = 0.599\%$
Total $= 0.021 + 0.198 + 0.599 \approx 0.82\%$
So the answer is (b) $0.82\%$ .

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