Explanation

(c) Before connecting the voltmeter, potential difference across $100\Omega$ resistance
$V_{1} = \frac{100}{(100 + 10)} \times V = \frac{10}{11}V$
Finally after connecting voltmeter across $100\Omega$ equivalent resistance
$\frac{100 \times 900}{(100 + 900)} = 90\Omega$
Final potential difference
$V_{f} = \frac{90}{(90 + 10)} \times V = \frac{9}{10}V$
% error $= \frac{V_{i} - V_{f}}{V_{i}} \times 100$
$= \frac{\frac{10}{11}V - \frac{9}{10}V}{\frac{10}{11}V} \times 100 = 1.0$