A 100ml solution having activity 50 dps is kept in a beaker It is now constantly diluted by adding water at a constant rate of `10ml//sec` and `2 ml//sec` of solution is constantly being taken out. Find the activity of 10 ml solution which is taken out, assuming half life to be effectively very large :

Explanation

To solve the problem, we need to find the activity of the 10 ml solution that is taken out from a beaker containing a radioactive solution. The initial conditions are as follows:

- Initial volume of the solution = 100 ml
- Initial activity of the solution (A₀) = 50 dps
- Rate of water being added = 10 ml/sec
- Rate of solution being taken out = 2 ml/sec
- We need to find the activity of the solution taken out after a certain time.

### Step-by-Step Solution:

1. **Determine the total volume of the solution at any time \( t \)**:
\[ V(t) = 100 \text{ ml} + (10 \text{ ml/sec} \times t) - (2 \text{ ml/sec} \times t) = 100 + 8t \text{ ml} \]

2. **Determine the activity of the solution at any time \( t \)**:
The activity of the solution is proportional to the number of radioactive nuclei present. Since we are assuming the half-life is very large, the activity will not change significantly over short periods. Therefore, we can use the initial activity \( A₀ \) and the dilution factor.

3. **Calculate the time to take out 10 ml of the solution**:
Since the solution is being taken out at a rate of 2 ml/sec, the time \( t \) to take out 10 ml is:
\[ t = \frac{10 \text{ ml}}{2 \text{ ml/sec}} = 5 \text{ seconds} \]

4. **Calculate the activity at \( t = 5 \) seconds**:
The activity at any time \( t \) can be derived from the formula:
\[ A(t) = A₀ \left( \frac{V₀}{V(t)} \right) \]
where \( V₀ \) is the initial volume (100 ml) and \( V(t) \) is the volume at time \( t \):
\[ V(5) = 100 + 8 \times 5 = 140 \text{ ml} \]
Thus, the activity at \( t = 5 \) seconds is:
\[ A(5) = 50 \left( \frac{100}{140} \right) = 50 \times \frac{5}{7} = \frac{250}{7} \text{ dps} \]

5. **Calculate the activity of the 10 ml solution taken out**:
The activity of the 10 ml solution taken out is given by:
\[ A_{out} = A(5) \times \frac{10 \text{ ml}}{V(5)} = \frac{250}{7} \times \frac{10}{140} \]
Simplifying this gives:
\[ A_{out} = \frac{2500}{980} = \frac{125}{49} \text{ dps} \]

### Final Answer:
The activity of the 10 ml solution taken out is approximately \( 2.55 \text{ dps} \).