As shown in the figure the voltmeter reads 2V across 5Ω resistor....

Let the resistance of the voltmeter be $RΩ$. The voltmeter is in parallel with the $5Ω$. So the total resistance in the circuit is

\left(\dfrac{5R}{5 + R} + 2\right)Ω$</p><p>Applying Ohm's law in the whole circuit,$V = IR$,</p><p>$I = \left(\dfrac{3}{2 + \dfrac{5R}{5 + R}}\right)A$</p><p>It is given that the potential difference across$5Ω$is$2V$.</p><p>Thus, by Ohm's law,</p><p>\left(\dfrac{3}{2 + \dfrac{5R}{5 + R}}\right) \times \left(\dfrac{5R}{R + 5}\right) = 2 ⇒ \dfrac{15 + 3R}{10 + 7R} \times \dfrac{5R}{5 + R} = 2 ⇒ 15R = 20 + 14R ⇒ R = 20Ω$