Please solve question number one. Send solution. Answer is option A.

Dear Student
Let on the plate 1, given chargeQdistributed evenly asQ2andQ2on the either side. Similarly, let on plate 2 charge2Qis distributed asQandQon either side.The charges need to be redistributed as: Two surfaces facing each other must have equal and opposite charges. Now induction will take place due to the higher chargeQthanQ2.
Therefore,−Qwill be induced on the right side of plate 1. This induced charge will make total charge on this side of plate 1 asQ2−Q=−Q2
This will lead to redistribution of charges on three electrically isolated plates before closing of the switch as shown in the figure attached


After closing of switch we see that plates 1 and 3 are at the same potential and plate 2 remains isolated. LetC1andC2becapacitanceof three plates as shown. Without loss of generality we can assume that charge of both faces of plate needs to redistributed. Therefore, charges on facing sides of plates 1 and 3 only move. Let the final charge distribution be as given in attached figure
Charge on isolated plate 2

q1+q2=2Q.......(1)

Potential between plates

V21=V23
⇒q1C1=q2C2.....(2)

NowC1=2C2, (due to separation being twice between plates.) With this we get

q1=2q2

Inserting this in (1) and solving we get

2q2+q2=2Q
q2=2Q3
Alsoq1=4Q3

Charge moved through switchK, movement from surface having more charge to surface having less charge is

−Q2−(−q1)
⇒−Q2+4Q3
⇒5Q6
Hope it helped