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

12.5 V

Voltage across terminal of a battery is $V = E - Ir = E - \frac{Er}{r+R} = \frac{ER}{r+R}$

$\Rightarrow \frac{ER}{r+R} = 5V$ ..............(1)

According to question $\frac{6ER}{r+6R} = 10$.......(2)

On dividing $\frac{6r+6R}{r+6R} = 2 \Rightarrow 6r+ 6R = 2r + 12R$

$ \Rightarrow r = 1.5R$

$ E = 12.5V$