A 100 ml solution having activity 50 dps is kept in a beaker. It is now constantly diluted by adding water at a constant rate of 10 ml/s and simultaneously 2 ml/s of solution is constantly taken out. The activity of 10 ml solution which is taken out, assuming half life to be effectively very large, is A0[1−(5a)1/b] where A0=50 dps. Find the value of ab.

A 100 ml solution having activity 50 dps is kept in a beaker. It is no

1.	A 100 ml solution having activity 50 dps is kept in a beaker. It is now constantly diluted by adding water at a constant rate of 10 ml/s and simultaneously 2 ml/s of solution is constantly taken out. The activity of 10 ml solution which is taken out, assuming half life to be effectively very large, is A0[1−(5a)1/b] where A0=50 dps. Find the value of ab.
Answer» A $100 ml$ solution having activity $50 dps$ is kept in a beaker. It is now constantly diluted by adding water at a constant rate of $10 ml/s$ and simultaneously $2 ml/s$ of solution is constantly taken out. The activity of $10 ml$ solution which is taken out, assuming half life to be effectively very large, is $A_{0} [1 - {(\frac{5}{a})}^{1 / b}]$ where $A_{0} = 50 dps$ . Find the value of $a b$ .