In the circuit shown, initially there is no charge on capacitors and keys S 1 and S 2 are open. The values…

For option $\left(\mathrm{C}\right)$ , refer circuit diagram below

Just after closing of switch charge on all the capacitors $\left(C_1,C_3\mathrm{a}\mathrm{n}\mathrm{d}{C}_4\right)=0$
$\therefore$ Replace all capacitors with wire, then circuit will get reduced as shown below.

$i=\frac{5}{70+100+30}=\frac{5}{200}=25mA\Rightarrow \left(\mathrm{C}\right)$ is correct
Now $S_1$ is kept closed for long time circuit is in steady state

$\Rightarrow$ Apply KVL in loop, start from battery
$+5-\frac{2}{80}-\frac{2}{10}-\frac{2}{80}=0$
$\Rightarrow \frac{10.2}{80}=5$
$\Rightarrow 2=40\mu C$
$\Rightarrow$ Potential difference across $C_1$
$\Delta V_{C_1}=\frac{2}{10}=\frac{40}{10}=4\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}$
$\Rightarrow \left(\mathrm{D}\right)$ is correct.
For option $\left(\mathrm{B}\right)\Rightarrow$ Now just after closing of $S_2$ charge on each capacitor remain same,
Refer circuit below -

KVL in left loop, starting from battery of $10$ volts,
$10-30x-\frac{40}{10}-70y=0$
$\Rightarrow 30x+70y=6$ ...(i)
Now, apply KVL in right loop starting from $80\mu F$ capacitor
$\Rightarrow -\frac{40}{80}+5+\left(x-y\right)30-\frac{40}{80}+\left(x-y\right)\times 100-10+x\times 30=0$
$\Rightarrow 160x-130y-6=0$ ...(ii)
Now, solving equation (i) and (ii) we get,
$y=\frac{96}{1510}$
and $x=0.05amp.$ $\Rightarrow \left(B\right)$ is incorrect.