In an LCR circuit, the resonating frequency is 500 kHz. If the value of L is increased two times and value of C is decreased `(1)/(8)` times, then the new resonating frequency in kHz will be -

Options

A. 250
B. 500
C. 1000
D. 2000

Answer

To solve the problem step by step, we will use the formula for the resonating frequency of an LCR circuit, which is given by:

\[ f_r = \frac{1}{2\pi\sqrt{LC}} \]

### Step 1: Write down the initial resonating frequency
The initial resonating frequency is given as:

\[ f_r = 500 \text{ kHz} \]

### Step 2: Identify the changes in L and C
According to the problem:
- The inductance \( L \) is increased to \( 2L \).
- The capacitance \( C \) is decreased to \( \frac{C}{8} \).

### Step 3: Substitute the new values into the frequency formula
The new resonating frequency \( f'_r \) can be expressed as:

\[ f'_r = \frac{1}{2\pi\sqrt{L' C'}} \]

Substituting the new values of \( L' \) and \( C' \):

\[ f'_r = \frac{1}{2\pi\sqrt{(2L) \left(\frac{C}{8}\right)}} \]

### Step 4: Simplify the expression
Now simplify the expression:

\[ f'_r = \frac{1}{2\pi\sqrt{\frac{2L \cdot C}{8}}} \]
\[ f'_r = \frac{1}{2\pi\sqrt{\frac{L \cdot C}{4}}} \]
\[ f'_r = \frac{1}{2\pi\left(\frac{1}{2}\sqrt{LC}\right)} \]
\[ f'_r = \frac{2}{1} \cdot \frac{1}{2\pi\sqrt{LC}} \]
\[ f'_r = 2 \cdot f_r \]

### Step 5: Substitute the value of the initial frequency
Now substitute the initial frequency \( f_r = 500 \text{ kHz} \):

\[ f'_r = 2 \cdot 500 \text{ kHz} = 1000 \text{ kHz} \]

### Final Answer
Thus, the new resonating frequency is:

\[ \boxed{1000 \text{ kHz}} \]

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