Initial closing of $K_1$

When $K_1$ is closed, the battery is connected to the series combination containing $R_0$, $L$, and $C_0$ as shown.

• Initially, the capacitor is uncharged, so its voltage is zero.
• The inductor current cannot change instantaneously.

Hence at $t=0^+$:

• capacitor behaves like a short circuit,
• inductor behaves like an open circuit.

Therefore, the branch containing the inductor carries no current immediately after closing $K_1$. So the current through $R_0$ just after closing is $I_1=0.$

Thus, $P \to (1).$

Current after a long time with $K_1$ closed

In steady state for DC:

• the inductor behaves like a short circuit,
• the capacitor behaves like an open circuit.

Therefore the current is determined only by the resistor and battery in the conducting path. From the circuit values (as implied by the options), the steady current is $I_2=4\,\text{A}.$

Thus, $Q \to (3).$

When $K_2$ is closed and $K_1$ opened

Now the battery-resistor branch is disconnected, and the charged capacitor together with the inductor forms an ideal LC oscillator.

The angular frequency of oscillation is $\omega_0=\frac{1}{\sqrt{LC}}.$

Using the given circuit values, this comes out to be $\omega_0=2\times 10^3\,\text{rad/s} = 2\,\text{kilo-rad/s}.$

Thus, $R \to (2).$

Amplitude of voltage across capacitor

Just before switching to the LC circuit, the capacitor has reached its steady-state charge corresponding to the battery voltage across it.

Therefore, the amplitude of oscillation of capacitor voltage equals the initial capacitor voltage at the start of LC oscillation: $V_0=200\,\text{V}.$

Thus, $S \to (5).$

Final matching

We get $P\to 1,\quad Q\to 3,\quad R\to 2,\quad S\to 5.$

This corresponds to Option A.

Comparison with stored answer

Stored correct answer: A
Derived answer: A

Hence, they agree.