Upon a six fold increase in the external resistance of a circuit, the voltage across the terminals of the battery has increased from 5 V to 10V. The e.m.f. of battery is

$\mathrm{E}-\frac{\mathrm{E}}{(\mathrm{R}+\mathrm{r})}\times \mathrm{r}=5$……………(1)
or $\mathrm{E}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{r}}\right)=5$……………..(2)
$\mathrm{E}-\frac{\mathrm{E}}{(6\mathrm{R}+\mathrm{r})}\times \mathrm{r}=10$………..(3)
or $\mathrm{E}\left(\frac{6\mathrm{R}}{6\mathrm{R}+\mathrm{r}}\right)=10$………….(4)
Dividing eq. (2) by eq. (4), we get
     2r= 3R
Putting in eq. (3). we get $\mathrm{E}-\frac{\mathrm{E}\times \mathrm{r}}{4\mathrm{r}+\mathrm{r}}=10$
Solving, we get  E = 12.5V