A capacitor of capacitance $\text{C=3}\mu \text{F}$ is first charged by connecting it across a 10V battery by closing key ${\text{K}}_1$. It is then allowed to get discharged through $2\Omega$  and $4\Omega$ resistor by closing the key ${\text{K}}_2$  and opening key ${\text{K}}_1$ . The total heat energy dissipated in the $2\Omega$ resistor is equal to
 
 

 
Before 1 K opened the energy stored over capacity $C_1=\frac{1}{2}C{V}^2$
$=\frac{1}{2}\times 3\times {10}^{-6}\times {10}^2=150\mu J$
2. Now this energy will be dissipated $(E=i^{2}Rt)$ in the two resistors in the ratio of $R_1:R_2$  as they are in series (current flowing through the resistors is same) when switch $K_2$ is closed.
$\begin{array}{l}\therefore 150\mu J\text{ is lost at the radio 2:4=1:2}\\ \text{In 2}\Omega \frac{1}{3}\text{rd of }150\mu J\text{and in 4}\Omega \frac{2}{3}\text{rd of }150\mu J\text{will be lost}\\ \text{Therefore, the correct answer is}\left(2\right)\end{array}$