In the given circuit, $R_1 = 6 Ω, R_2 = 3Ω, C_1 = 4 μF$ and $C_2 = 5 μF$. When key K is closed, the charge stored in $C_1$ is

24 μC

The correct answer is Option (1) → 24 μC

Given: $R_1=6\Omega,\;R_2=3\Omega,\;C_1=4\mu\text{F},\;C_2=5\mu\text{F},\;V=9\text{ V}$

When key $K$ is closed and steady state is reached, capacitors are open circuits and node voltage at junction of $R_1$ and $R_2$ is given by voltage divider:

$V_{\text{node}}=V\cdot\frac{R_2}{R_1+R_2}=9\cdot\frac{3}{6+3}=9\cdot\frac{1}{3}=3\text{ V}$

If $C_1$ is connected between the battery positive (9 V) and that node, voltage across $C_1$ is

$V_{C_1}=9-3=6\text{ V}$

Charge on $C_1$:

$Q_1=C_1V_{C_1}=4\mu\text{F}\times 6\text{ V}=24\mu\text{C}$

Answer: $24\mu\text{C}$