Q 11-01-144JEE MainJEE Main 2023 (8 Apr, Shift 1)Easy
A cylindrical wire of mass $(0.4\pm0.01)$ g has length $(8\pm0.04)$ cm and radius $(6\pm0.03)$ mm. The maximum error in its density will be
Answer: (D) $4\%$
$\rho=\dfrac{m}{\pi r^2l}$, so
$$\frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+2\frac{\Delta r}{r}+\frac{\Delta l}{l}=\frac{0.01}{0.4}+2\times\frac{0.03}{6}+\frac{0.04}{8}=2.5\%+1\%+0.5\%=4\%$$
Solution by Sreeraj P, M.Sc Physics