Q 11-01-076JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
The maximum percentage error in the measurement of density of a wire is [Given, mass of wire $= (0.60\pm0.003)\ \text{g}$, radius of wire $= (0.50\pm0.01)\ \text{cm}$, length of wire $= (10.00\pm0.05)\ \text{cm}$]
Answer: (B) $5$
$\rho = \dfrac{m}{\pi r^2 l}$, so
$$\frac{\Delta\rho}{\rho}\times100 = \left(\frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}\right)\times100$$
$$= \frac{0.003}{0.60}\times100 + 2\times\frac{0.01}{0.50}\times100 + \frac{0.05}{10.00}\times100 = 0.5 + 4 + 0.5 = 5\%$$
Solution by Sreeraj P, M.Sc Physics