A potential difference $\mathrm{V}=100 \pm 5\mathrm{V}$, when applied across a resistance, gives a current $\mathrm{I}=10 \pm 0.2 \mathrm{A}$. What is the percentage error in $\mathrm{R}$?

Potential difference is $\mathrm{V}=100 \pm 5\mathrm{V}$

Current is $\mathrm{I}=10 \pm 0.2\mathrm{A}$.

Percentage error in voltage is obtained as

$\frac{∆\mathrm{V}}{\mathrm{V}}=\frac{5}{100}\times 100\%=5\%$

Percentage error in current is obtained as 

$\frac{∆\mathrm{I}}{\mathrm{I}}=\frac{0.2}{10}\times 100\%=2\%$

Using, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$, the percentage error in resistance is calculated as $\mathrm{R}=5\%+2\%=7\%$.