The resistance $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{i}}$  where   $\mathrm{V}=\left(100\pm 5\right) \mathrm{volt}\mathrm{and} \mathrm{i}=\left(10\pm 0.2\right)\mathrm{Amp}.$ Then the maximum error in ‘R’ is ______

$\begin{array}{l}\mathrm{Base}\mathrm{value}\mathrm{of}\mathrm{R}=\frac{100}{10}=10\mathrm{\Omega }.\\ \mathrm{Now},\mathrm{R}={\mathrm{Vi}}^{-1}\\ \therefore {\left(\frac{\mathrm{\Delta R}}{\mathrm{R}}\times 100\right)}_{\max }\mathrm{\%}=\left(\frac{\mathrm{\Delta V}}{\mathrm{V}}\times 100\right)\mathrm{\%}+\left(\frac{\mathrm{\Delta i}}{\mathrm{i}}\times 100\right)\times 100\mathrm{\%}\\ =\frac{5}{100}\times 100\mathrm{\%}+\frac{0.2}{10}\times 100\mathrm{\%}=7\mathrm{\%}\end{array}$