In the circuit shown, when switch S₁ is closed and S₂ is open, the ideal voltmeter shows a reading of 18 V. When switch S₂ is closed and S₁ is open, the reading of voltmeter is 24 V. When S₁ and S₂ both are closed, the voltmeter reading will be

Terminal potential difference across the terminals of cell E

$\mathrm{V}=\mathrm{E}-\mathrm{ir}=\mathrm{E}-\left(\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}}\right)\mathrm{r}=\frac{\mathrm{ER}}{\mathrm{R}+\mathrm{r}}$

According to question,

$\begin{array}{l}18=\frac{6\mathrm{E}}{6+\mathrm{r}}\\ \mathrm{and}24=\frac{12\mathrm{E}}{12+\mathrm{r}}\end{array}$

Solving (i) and (ii), we get $r=6\Omega ,E=36V$
When both S₁ and S₂ are closed, the effective external resistance in the circuit will be,

$\begin{array}{l}\mathrm{R}=\frac{6\times 12}{18}=4\mathrm{\Omega }\\ \mathrm{So},\mathrm{V}=\frac{\mathrm{ER}}{\mathrm{r}+\mathrm{R}}=\frac{36\times 4}{6+4}=\frac{144}{10}\end{array}$

Thus, choice (1) is correct.