Question 90: Units , Dimensions and Error Analysis

Question

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:

Accepted Answer

$$\begin{aligned}
& \mathrm{R}=\frac{ho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \
& \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% 	ext { and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \%
\end{aligned}$$

Answer

0.144%

Answer

0.2%

Answer

0.1%

Answer

0.3%

Units , Dimensions and Error Analysis

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:

Choose the correct answer: