The reading in the ideal voltmeter (V) shown in the given circuit...

As the voltmeter is idea, no current will flow through it. Therefore, considering left loop, we can write

$i = \dfrac{{E}_{eq}}{{r}_{eq}} = \dfrac{8 \times 5}{8 \times 0 .2}$

$⇒ I = 25A$

Now, potential drop across the cell which is connected in parallel with Voltmeter,

$V = E - ir$

$= 5 - 0 . 2 \times 25$

$= 0V$