The emf of a battery is 2V and its internal resistance is `0.5Omega` the maximum power which it can deliver to any external circuit will be

Explanation

To find the maximum power that a battery can deliver to an external circuit, we can use the formula for maximum power transfer, which states:

\[ P_{\text{max}} = \frac{E^2}{4r} \]

where:
- \( P_{\text{max}} \) is the maximum power,
- \( E \) is the electromotive force (emf) of the battery,
- \( r \) is the internal resistance of the battery.

### Step-by-Step Solution

1. **Identify the Given Values:**
- The emf of the battery, \( E = 2 \, \text{V} \)
- The internal resistance, \( r = 0.5 \, \Omega \)

2. **Substitute the Values into the Formula:**
\[ P_{\text{max}} = \frac{E^2}{4r} \]
Substituting the values we have:
\[ P_{\text{max}} = \frac{(2 \, \text{V})^2}{4 \times 0.5 \, \Omega} \]

3. **Calculate \( E^2 \):**
\[ (2 \, \text{V})^2 = 4 \, \text{V}^2 \]

4. **Calculate the Denominator:**
\[ 4 \times 0.5 \, \Omega = 2 \, \Omega \]

5. **Calculate \( P_{\text{max}} \):**
\[ P_{\text{max}} = \frac{4 \, \text{V}^2}{2 \, \Omega} = 2 \, \text{W} \]

6. **Conclusion:**
The maximum power that the battery can deliver to any external circuit is \( 2 \, \text{W} \).

### Final Answer:
The maximum power that can be delivered to any external circuit is **2 Watts**.