Ten identical cells each of potential E and internal resistance r...

Given: Initial number of cells $\left(n_1\right)=10$; Potential of each cell $=E$; Internal resistance of each cell $=r and final number of cells \left(n_2\right)=3$. We know from the Ohm's law, total voltage of ten cells $=10 \times E=10 E$ and total resistance in ten cells $=10 \times r=10 r$. Therefore current in the circuit, $I=\frac{10 E}{10 r}=\frac{E}{r}$ or potential difference across three cells $=I \times 3 r=\frac{E}{r} \times 3 r=3 E$. (Since the voltmeter is ideal, therefore it will $\operatorname{read} 3 E)$.