A 10V battery with internal resistance $1\Omega$  and a 15V battery with internal resistance 0.6$\Omega$ are Connected in parallel to a voltmeter (see figure).The reading in the voltmeter will be close to in volts.

As the two cells oppose each other hence, the effective emf in closed circuit is 15-10=5V and net resistance is 1+0.6= $1.6\Omega$   (because in the closed circuit the internal resistance of two cells are in series)
current  in the circuit=I= effective emf /  total resistance = 5/1.6 A
The potential difference across voltmeter will be same as the terminal voltage of either cell
Since  the current I is drawn from the cell  of 15V 
$\begin{array}{l}\therefore V_1=E_1-I{r}_{\begin{array}{l}1\\ \end{array}}\\ =15-\frac{5}{1.6}\times 0.6=13.125V\end{array}$