A pump can fill a tank with water in $2$ hours. Because of a leak developed in the tank, it now takes $2.5$ hours to fill the tank. In how many hours will the leak empty the full tank when the pump is not running?

Rate of the pump working alone = $\frac{1}{2}$ tank per hour.
Rate of the pump with the leak operating simultaneously = $\frac{1}{2.5}=\frac{2}{5}$ tank per hour.

Let the emptying rate of the leak be $\frac{1}{L}$ tank per hour.

$$\frac{1}{2}-\frac{1}{L}=\frac{2}{5}$$

$$\frac{1}{L}=\frac{1}{2}-\frac{2}{5}$$

Taking LCM of $2$ and $5$, which is $10$:

$$\frac{1}{L}=\frac{5-4}{10}=\frac{1}{10}$$

$$L=10\text{ hours}$$

Therefore, the leak will completely empty a full tank in $10$ hours.