The potential difference across the $100\Omega$ resistance in the following circuit is measured by a voltmeter of $900\Omega$ resistance. The percentage error made in reading the potential difference is

Before connecting the voltmeter, potential difference across $100\mathrm{\Omega }$

${\mathrm{V}}_{\mathrm{i}}=\frac{100}{\left(100+10\right)}\times \mathrm{V}=\frac{10}{11}\mathrm{V}$

Finally after connecting voltmeter across $100\Omega$

Equivalent resistance $\frac{100\times 900}{\left(100+900\right)}90\mathrm{\Omega }$

Final potential difference,

$\begin{array}{l}{\mathrm{V}}_{\mathrm{f}}=\frac{90}{\left(90+10\right)}\times \mathrm{V}=\frac{9}{10}\mathrm{V}\\ \mathrm{\%}\mathrm{error}=\frac{{\mathrm{V}}_{\mathrm{i}}-{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\times 100\\ =\frac{\frac{10}{11}\mathrm{V}-\frac{9}{10}\mathrm{V}}{\frac{10}{11}\mathrm{V}}\times 100=1.0\mathrm{\%}\end{array}$