A  $2\mu F$  capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch  $S$  is turned to position 2 is

Initially and when switch  $S$  is turned to position -2 charge $\left(q\right)$  will remain constant.  Initially energy stored
 $U_i=\frac{1}{2}\frac{q^2}{C_i}=\frac{1}{2}\times \frac{q^2}{2}=\frac{q^2}{4}$
When switch  $S$  is turned to position- $2$  then energy stored.
$U_f=\frac{1}{2}\frac{q^2}{C_f}=\frac{1}{2}\times \frac{q^2}{(2+8)}=\frac{q^2}{20}$ 
$\therefore$  Energy dissipated,  $△U=U_i-U_f=\frac{q^2}{4}-\frac{q^2}{20}=\frac{q^2}{5}$
 $\therefore \%$  of stored energy dissipated
 $=\frac{△U}{U_i}\times 100=\frac{q^2}{5}\times \frac{4}{q^2}\times 100=80\%$