Question

An ideal inductor is connected across a capacitor. Oscillations of energy \(K\) are set up in the circuit. The capacitor plates are slowly drawn apart such that the frequency of oscillations is quadrupled. The work done in the process is

• A. \(15K\)
• B. \(13K\)
• C. zero
• D. \(2K\)

Hint:

In an LC circuit, \(f \propto \frac{1}{\sqrt{C}}\). If frequency becomes 4 times, capacitance becomes \(\frac{1}{16}\) times and energy becomes 16 times.

Answer 1

The Correct Option is A

Step 1: Frequency of LC oscillation.
For an LC circuit, frequency of oscillation is:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

Step 2: Relate frequency with capacitance.
Since \(L\) is constant, frequency varies with capacitance as:
\[ f \propto \frac{1}{\sqrt{C}} \]

Step 3: Frequency is quadrupled.
Given final frequency is four times initial frequency:
\[ f' = 4f \]
Therefore, capacitance must become:
\[ C' = \frac{C}{16} \]

Step 4: Energy stored in capacitor.
Energy stored in capacitor is:
\[ U = \frac{q^2}{2C} \]
When plates are slowly separated, charge remains same and capacitance decreases.

Step 5: Calculate final energy.
Initial energy is \(K\). Since \(C' = \frac{C}{16}\), energy becomes 16 times.
\[ K' = 16K \]

Step 6: Work done in the process.
Work done is equal to increase in energy.
\[ W = K' - K \]
\[ W = 16K - K = 15K \]
\[ \boxed{15K} \] Hence, correct answer is option (A).