JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 16)
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:
0.144%
0.2%
0.1%
0.3%
Explanation
$$\begin{aligned}
& \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \\
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \\
& \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% \text { and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \\
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \%
\end{aligned}$$
Comments (0)
