Q.

In the circuit shown (figure1), all the switches are open initially and all capacitors are uncharged. Now first only  S1 is closed and open when steady state is reached, then  S2 is closed and open when again steady state is reached, after this S3  is closed and open when steady state is reached, this process repeated continuously. After closing sixth switch, the fourth capacitor (8C) is removed and connected (say at t = 0) to an uncharged capacitor of original capacitance C which now has a dielectric (dielectric constant 2) filling half its volume (see figure2), through a resistor R.

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a

Final charge on capacitor B when steady state reached in figure 2  12823CV

b

Final charge (when steady state reached in figure 2) on capacitor of capacitance 8C is  12857CV

c

Final charge (when steady state reached in figure 2) on capacitor of capacitance 8C is 8119CV

d

Final charge on capacitor B when steady state reached in figure 2 819CV

answer is B, C.

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Detailed Solution

After closing S1 charge on C = 27 CV
After closing S2 charge on  2C=2C.27CV3C=18CV
After closing S3 charge on 4C=4C.18CV6C=12CV
After closing S4  charge on   8C=8C.12CV12C=8CV
After closing S5 charge on 8C=8C.8CV24C=8CV3
Final charge on 8C in figure (2)
 Q=8C.8CV38C+3C2=64CV3192=128CV57
Final charge on capacitor B is
 Q=3C2.8CV38C+3C2=8CV19
E at P=ΔVd=16V57d

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