Answer:
Option A,B,C,D
Explanation:

Just after pressing key,
5-25000i1 =0
5-50000 i2 =0
( As charge in both capacitors =0)
⇒i1=0.2mA⇒i2=0.1mA
And VB+25000i1=VA
⇒VB−VA=−5V
After a long time , i1 and i2 =0 ( steady state)
⇒5−q140=0⇒q1=200μC
and 5−q220=0⇒q2=100μC
VB−q220=VA⇒VB−VA=+5V
⇒ (a) is correct.
For capacitor 1,
q1=200[1−e−tl]μC⇒i1=15e−tlmA
For capacitor 2,
q2=100[1−e−tl]μC⇒i2=110e−tlmA
⇒VB−q220+i1×25=VA
⇒VB−VA=5[1−e−t]−5e−t
=−5[1−2e−t]
At t=ln 2,
VB−V=5[1−1]=0
⇒ (b) is correct.
At t=1,
i=i1+i2=15e−1+110e−1=310.1e
At t=0
i=i1+i2=15+110=310
⇒ (c) is correct
After a long time , i1=i2=0
⇒ (d) is correct.