Correct Answer - A
a. Before closing the switch After closing the switch
`q_(1)+q_(2)=2Q`
`V_(AB)=V_(AC)` or `(Q_(1))/(C_(1))=(q_(2))/(C_(2))`
`(q_(1))/(q_(2))=(C_(1))/(C_(2))=2` or `q_(1)=2q_(2)`
Solving, we get `q_(1)=(4Q)/(3), q_(2)=(2Q)/(3)`
Charge flown through `K` is
`-(Q)/(2)-(-q_(1))- -(Q)/(2)+(4Q)/(3)=5Q//6`