A `2 mu F` condenser is charged to `500 V` and then the plates are joined through a resistance. The heat produced in the resistance is joule is.

Options

A. `50 xx 10^(-2)` Joule
B. `25 xx 10^(-2)` Joule
C. `0.25 xx 10^(-2)` Joule
D. `0.5 xx 10^(-2)` Joule

Explanation

To find the heat produced in the resistance when a charged capacitor is discharged, we can use the formula for the energy stored in a capacitor. The energy (U) stored in a capacitor is given by the formula:

\[ U = \frac{1}{2} C V^2 \]

where:
- \( U \) is the energy in Joules,
- \( C \) is the capacitance in Farads,
- \( V \) is the voltage in Volts.

### Step-by-step Solution:

1. **Identify the values:**
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Voltage \( V = 500 \, V \)

2. **Substitute the values into the energy formula:**
\[ U = \frac{1}{2} C V^2 = \frac{1}{2} (2 \times 10^{-6}) (500)^2 \]

3. **Calculate \( V^2 \):**
\[ V^2 = 500^2 = 250000 \]

4. **Substitute \( V^2 \) back into the energy formula:**
\[ U = \frac{1}{2} (2 \times 10^{-6}) (250000) \]

5. **Calculate the product:**
\[ U = (1 \times 10^{-6}) (250000) = 0.25 \, J \]

6. **Final result:**
The heat produced in the resistance when the capacitor is discharged is:
\[ U = 0.25 \, J \]

### Final Answer:
The heat produced in the resistance is **0.25 Joules**.