At a certain frequency `omega_(1)`, the reactance of a certain capacitor equals that of a certain inductor. If the frequency is changed to`omega_(2) = 2omega_(1)`, the raito of reactance of the inductor to that of the capacitor is :

Options

A. `4:1`
B. `sqrt2:1`
C. `1:2sqrt2`
D. `1:2`

Answer

To solve the problem, we need to find the ratio of the reactance of the inductor (XL) to the reactance of the capacitor (XC) when the frequency is changed from ω1 to ω2 = 2ω1.

### Step-by-Step Solution:

1. **Understand the Reactance of Inductor and Capacitor**:
- The reactance of an inductor (XL) is given by:
\[ X_L = \omega L \]
- The reactance of a capacitor (XC) is given by:
\[ X_C = \frac{1}{\omega C} \]

2. **Initial Condition**:
- At frequency ω1, it is given that:
\[ X_L = X_C \]
- Therefore, we can write:
\[ \omega_1 L = \frac{1}{\omega_1 C} \]
- Rearranging gives:
\[ \omega_1^2 LC = 1 \quad \text{(1)} \]

3. **Change of Frequency**:
- Now, we change the frequency to ω2 = 2ω1. We need to find the new reactances:
\[ X_L' = \omega_2 L = 2\omega_1 L \]
\[ X_C' = \frac{1}{\omega_2 C} = \frac{1}{2\omega_1 C} \]

4. **Calculate the Ratio of Reactances**:
- Now, we find the ratio of the new reactance of the inductor to that of the capacitor:
\[ \frac{X_L'}{X_C'} = \frac{2\omega_1 L}{\frac{1}{2\omega_1 C}} = 2\omega_1 L \cdot 2\omega_1 C = 4\omega_1^2 LC \]

5. **Substituting from Equation (1)**:
- From equation (1), we know that:
\[ \omega_1^2 LC = 1 \]
- Therefore:
\[ \frac{X_L'}{X_C'} = 4 \cdot 1 = 4 \]

6. **Final Answer**:
- The ratio of the reactance of the inductor to that of the capacitor at frequency ω2 is:
\[ \frac{X_L'}{X_C'} = 4 \]

### Conclusion:
The ratio of the reactance of the inductor to that of the capacitor when the frequency is changed to ω2 = 2ω1 is **4**.