An inductance of 1 mH a condenser of `10 mu F` and a resistance of `50 Omega` are connected in series. The reactances of inductor and condensers are same. The reactance of either of them will be

Options

A. `100 Omega`
B. `30 Omega`
C. `3.2 Omega`
D. `10 Omega`

Explanation

To solve the problem, we need to find the reactance of either the inductor or the capacitor when they are connected in series and their reactances are equal. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Inductance (L) = 1 mH = \(1 \times 10^{-3}\) H
- Capacitance (C) = 10 µF = \(10 \times 10^{-6}\) F
- Resistance (R) = 50 Ω

### Step 2: Write the formulas for inductive and capacitive reactance
- Inductive reactance (\(X_L\)) is given by the formula:
\[ X_L = \omega L \]
- Capacitive reactance (\(X_C\)) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]

### Step 3: Set the reactances equal to each other
Since it is given that the reactances are the same:
\[ X_L = X_C \]
This leads to:
\[ \omega L = \frac{1}{\omega C} \]

### Step 4: Rearrange the equation to find \(\omega\)
Multiplying both sides by \(\omega\):
\[ \omega^2 L = 1/C \]
Thus,
\[ \omega^2 = \frac{1}{LC} \]
Taking the square root:
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 5: Substitute the values of L and C
Now substituting the values of L and C:
\[ \omega = \frac{1}{\sqrt{(1 \times 10^{-3})(10 \times 10^{-6})}} \]

### Step 6: Calculate the value of \(\omega\)
Calculating the product:
\[ LC = (1 \times 10^{-3})(10 \times 10^{-6}) = 10 \times 10^{-9} = 10^{-8} \]
Thus,
\[ \omega = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10^4 \text{ rad/s} \]

### Step 7: Calculate the reactance
Now, we can calculate the inductive reactance \(X_L\):
\[ X_L = \omega L = (10^4)(1 \times 10^{-3}) = 10 \, \Omega \]

### Conclusion
The reactance of either the inductor or the capacitor is:
\[ \boxed{10 \, \Omega} \]