The correct option is
D By closing the switch the capacitance of the system increases
When the switch
S is open, no charge is present on inner shell.
Vinner shell=Vouter shell=kQ3R
So, Option
(a) is correct
Let the charge on the inner shell after closing the switch be
Q′
Since the inner shell is earthed, the net potential must be zero for it.
Vinner=0
The net potential at the inner shell is the sum of potential due to the inner shell itself and due to the outer shell.
⇒kQ′R+kQ3R=0⇒Q′=−Q3
So, Option
(b) and
(c) are correct.
Now from the definition of capacitance,
Ci=4πε0(3R)
The charge on inner shell is zero initially when switch is open, hence only capacitance of outer sphere is taken into account.
& Cf=4πε0(3R)+4πϵ0(1R−13R)
After the switch is closed the capacitance of both spheres add up, since they get connected in parallel.
⇒Cf>Ci
Hence, options (a) , (b) , (c) and (d) are the correct answers.