The density of a uniform cylinder is determined by measuring its mass , length and diameter . The measured values of

Question

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 ______. [2026]

Smart Achievers · Accepted Answer

(2)

$ρ = \frac{m}{π r^{2} l} = \frac{m}{\frac{π d^{2}}{4} l} = \frac{4 m}{π d^{2} l}$

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

$% error = (\frac{0.02}{97.42} + \frac{2 \times 0.02}{20.20} + \frac{0.05}{8.35}) \times 100 = 0.82 %$

Solution

Q.
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 ______. [2026]

1 0.63%

2 0.82%

3 0.72%

4 0.25%

Ans.
(2)

$ρ = \frac{m}{π r^{2} l} = \frac{m}{\frac{π d^{2}}{4} l} = \frac{4 m}{π d^{2} l}$

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

$% error = (\frac{0.02}{97.42} + \frac{2 \times 0.02}{20.20} + \frac{0.05}{8.35}) \times 100 = 0.82 %$

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