The natural frequency of an `LC`- circuit is 1,25,000 cycles per second. Then the capacitor `C` is replaced by another capacitor with a dielectric medium of dielectric constant k. In this case, the frequency decreases by 25 kHz. The value of k is

Options

A. 3.05
B. 2.18
C. 1.56
D. 1.74

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- The initial frequency \( f_1 = 125,000 \) cycles per second (or \( 125 \) kHz).
- The decrease in frequency due to the introduction of the dielectric is \( 25 \) kHz.
- Therefore, the new frequency \( f_2 = f_1 - 25 \) kHz = \( 125 \) kHz - \( 25 \) kHz = \( 100 \) kHz.

### Step 2: Write the relationship between frequency and capacitance
The natural frequency \( f \) of an LC circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
From this, we can derive that:
\[ f \propto \frac{1}{\sqrt{C}} \]
This indicates that the frequency is inversely proportional to the square root of the capacitance.

### Step 3: Set up the ratio of frequencies
Using the relationship derived above, we can set up the following ratio:
\[ \frac{f_1}{f_2} = \sqrt{\frac{C_2}{C_1}} \]
Where \( C_1 \) is the capacitance without the dielectric and \( C_2 \) is the capacitance with the dielectric.

### Step 4: Substitute the known values
Substituting the values of \( f_1 \) and \( f_2 \):
\[ \frac{125}{100} = \sqrt{\frac{C_2}{C_1}} \]
This simplifies to:
\[ \frac{5}{4} = \sqrt{\frac{C_2}{C_1}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{5}{4}\right)^2 = \frac{C_2}{C_1} \]
Calculating the left side:
\[ \frac{25}{16} = \frac{C_2}{C_1} \]

### Step 6: Relate capacitance with dielectric constant
We know that the capacitance with a dielectric is given by:
\[ C_2 = k \cdot C_1 \]
Thus, we can express the ratio as:
\[ \frac{C_2}{C_1} = k \]
From our previous calculation, we have:
\[ k = \frac{C_2}{C_1} = \frac{25}{16} \]

### Step 7: Calculate the value of \( k \)
Now, we can calculate the value of \( k \):
\[ k = 1.5625 \]

### Conclusion
Thus, the value of the dielectric constant \( k \) is approximately \( 1.56 \).

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