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Q.

In the arrangement shown, when a dielectric slab having dielectric constant K is inserted in the capaciter of capacitance C, the reading of the volt meter becomes 1/5th of previous value.Then value of K is

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a

3

b

4

c

2

d

3.5

answer is B.

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Detailed Solution

Initial reading, V=2C3.EC=2E3Q1 = final charge on the capacitor = CK×2C(CK+2C)×E=2KCE(K+2)∴V' =Final potential = Q1CK=2EK+2Since V' = ½ V we can write 2EK+2=12×2E3⇒K=4
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