Q 11-01-162JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
A silver wire has a mass $(0.6 \pm 0.006)$ g, radius $(0.5 \pm 0.005)$ mm and length $(4 \pm 0.04)$ cm. The maximum percentage error in the measurement of its density will be
Answer: (C) $4\%$
Density $\rho = \dfrac{m}{\pi r^2 L}$, so
$$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$$
Each relative error is $1\%$: $\dfrac{0.006}{0.6} = \dfrac{0.005}{0.5} = \dfrac{0.04}{4} = 0.01$.
$$\frac{\Delta\rho}{\rho}\times 100 = 1\% + 2(1\%) + 1\% = 4\%$$
Solution by Sreeraj P, M.Sc Physics