The circuit is closed for long time so that the capacitors are fully charged. Switch is opened at a certain instant. The heat developed in resistor $R_2$  after that instant is $K\times {10}^{-4}J.$ Then find the value of K.

$C_{eq}=3\mu F$

P.D.   across $R_1$ in steady state =  $\frac{30}{100+50}\times 100=20V$
$\therefore$   Energy  stored $=\frac{1}{2}{C}_1{V}^2+\frac{1}{2}{C}_2{V}^2$
$U=\frac{1}{2}(1+2)\times {10}^{-6}\times 400=6\times {10}^{-4}J$
After switch is opened, heat is generated in $R_1$  and $R_2$ only.
Heat developed, 
$$\begin{gathered} \frac{H_1}{H_2}=\frac{{\displaystyle \int I^2{R}_{1}dt}}{I^2{R}_{2}dt}=\frac{R_1}{R_2} \\ {H}_2=\frac{U{R}_2}{R_1+R_2}=6\times {10}^{-4}\times \frac{200}{300}=4\times {10}^{4}J \end{gathered}$$