Q 11-01-088JEE MainJEE Main 2024 (1 Feb, Shift 1)Medium
The radius $r$, length $l$ and resistance $R$ of a metal wire were measured in the laboratory as
$$r = 0.35 \pm 0.05\ \text{cm},\quad R = 100 \pm 10\ \Omega,\quad l = 15 \pm 0.2\ \text{cm}$$
The percentage error in the resistivity of the material of the wire is
Answer: (B) $39.9\%$
Resistivity $\rho = \dfrac{R\pi r^2}{l}$, so the fractional errors add:
$$\frac{\Delta\rho}{\rho}\times100 = \frac{\Delta R}{R}\times100 + 2\frac{\Delta r}{r}\times100 + \frac{\Delta l}{l}\times100$$
$$= \frac{10}{100}\times100 + 2\times\frac{0.05}{0.35}\times100 + \frac{0.2}{15}\times100 = 10 + 28.57 + 1.33 \approx 39.9\%$$
Solution by Sreeraj P, M.Sc Physics