Q 11-01-122JEE MainJEE Main 2024 (31 Jan, Shift 1)Easy
If the percentage errors in measuring the length and the diameter of a wire are $0.1\%$ each, the percentage error in measuring its resistance will be:
Answer: (B) $0.3\%$
$R = \dfrac{\rho L}{\pi d^2/4}$, so
$$\frac{\Delta R}{R}\times100 = \frac{\Delta L}{L}\times100 + 2\frac{\Delta d}{d}\times100 = 0.1 + 0.2 = 0.3\%$$
Solution by Sreeraj P, M.Sc Physics