Q 11-01-052JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
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 $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is ______.
Answer: (B) $0.82\%$
$\rho = \dfrac{m}{\pi d^2 l/4}$, so
$$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + 2\frac{\Delta d}{d} = \frac{0.02}{97.42} + \frac{0.05}{8.35} + 2 \times \frac{0.02}{20.20}$$
$= 0.00021 + 0.00599 + 0.00198 = 0.00817$, i.e. about $0.82\%$.
Solution by Sreeraj P, M.Sc Physics