The frequency for which a 5`mu`F capacitor has a reactance of `1/1000 Omega`is given by
A.`100/pi` MHz B.`1000/pi` MHz C.`1/1000 Hz` D.1000 Hz

Options

A. `100/pi` MHz
B. `1000/pi` MHz
C. `1/1000 Hz`
D. 1000 Hz

Explanation

To solve the problem, we need to find the frequency at which a 5 µF capacitor has a reactance of 1/1000 Ω. We'll use the formula for capacitive reactance and the relationship between frequency and angular frequency.

### Step-by-step Solution:

1. **Identify Given Values**:
- Capacitance, \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- Capacitive reactance, \( X_c = \frac{1}{1000} \, \Omega \)

2. **Use the Formula for Capacitive Reactance**:
The formula for capacitive reactance is given by:
\[ X_c = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) (angular frequency) and \( f \) is the frequency in Hz.

3. **Rearranging the Formula**:
From the formula for capacitive reactance, we can rearrange it to find \( \omega C \):
\[ \omega C = 1000 \]

4. **Substituting for \( \omega \)**:
Substitute \( \omega = 2\pi f \) into the equation:
\[ 2\pi f C = 1000 \]

5. **Solve for Frequency \( f \)**:
Rearranging the equation gives:
\[ f = \frac{1000}{2\pi C} \]

6. **Substituting the Value of \( C \)**:
Substitute \( C = 5 \times 10^{-6} \, F \):
\[ f = \frac{1000}{2\pi \times 5 \times 10^{-6}} \]

7. **Calculating the Frequency**:
Simplifying the expression:
\[ f = \frac{1000}{10\pi \times 10^{-6}} = \frac{100 \times 10^6}{\pi} \]
Thus,
\[ f = \frac{100}{\pi} \, \text{MHz} \]

8. **Final Answer**:
The frequency for which the 5 µF capacitor has a reactance of \( \frac{1}{1000} \, \Omega \) is:
\[ \boxed{\frac{100}{\pi} \, \text{MHz}} \]