Explanation

(d) Resistance between $A$ and $B = \frac{1000 \times 500}{(1500)} = \frac{1000}{3}$
So, equivalent resistance of the circuit
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$R_{eq} = 500 + \frac{1000}{3} = \frac{2500}{3}$
$\therefore$ Current drawn from the cell
$i = \frac{10}{(2500/3)}\ = \frac{3}{250}A$
Reading of voltmeter $i.e.$
Potential difference across $AB = \frac{3}{250} \times \frac{1000}{3} = 4\ V$