The capacitor of an oscillatory circuit of frequency `10000 Hz` is enclosed in a container. When the container is evacuated, the frequency changes by`50 Hz`, the delectric constant of the gas is

Options

A. `1.1`
B. `1.01`
C. `1.001`
D. `1.0001`

Explanation

To solve the problem step by step, we will use the relationship between the frequency of an oscillatory circuit and the capacitance and inductance involved, particularly focusing on the effect of the dielectric constant.

### Step-by-Step Solution:

1. **Understand the Frequency Formula**:
The frequency \( f \) of an oscillatory circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance.

2. **Set Up the Initial Conditions**:
Initially, the frequency is given as \( f = 10,000 \, \text{Hz} \).

3. **Consider the Change in Frequency**:
When the container is evacuated, the frequency changes to \( f + 50 \, \text{Hz} \). Thus, the new frequency can be expressed as:
\[ f + 50 = \frac{1}{2\pi\sqrt{\frac{LC}{K}}} \]
where \( K \) is the dielectric constant of the gas.

4. **Write the Two Frequency Equations**:
- For the initial frequency:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
- For the frequency after evacuation:
\[ f + 50 = \frac{1}{2\pi\sqrt{\frac{LC}{K}}} \]

5. **Divide the Two Equations**:
Dividing the second equation by the first gives:
\[ \frac{f + 50}{f} = \frac{\frac{1}{2\pi\sqrt{\frac{LC}{K}}}}{\frac{1}{2\pi\sqrt{LC}}} \]
This simplifies to:
\[ \frac{f + 50}{f} = \sqrt{K} \]

6. **Square Both Sides**:
Squaring both sides results in:
\[ \left(\frac{f + 50}{f}\right)^2 = K \]
Expanding the left side gives:
\[ K = \frac{(f + 50)^2}{f^2} \]

7. **Substitute the Value of Frequency**:
Substitute \( f = 10,000 \):
\[ K = \frac{(10,000 + 50)^2}{(10,000)^2} = \frac{(10,050)^2}{(10,000)^2} \]

8. **Calculate the Value**:
Calculate \( (10,050)^2 \):
\[ (10,050)^2 = 101,002,500 \]
Calculate \( (10,000)^2 \):
\[ (10,000)^2 = 100,000,000 \]
Thus:
\[ K = \frac{101,002,500}{100,000,000} = 1.010025 \]

9. **Approximate the Dielectric Constant**:
Since \( K \) is very close to 1, we can approximate:
\[ K \approx 1 + \frac{50}{10,000} = 1 + 0.005 = 1.01 \]

### Final Answer:
The dielectric constant \( K \) of the gas is approximately \( 1.01 \).