The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit.

Two identical air capacitors of capacitance C are connected to a battery of EMF V and initially switch is closed for a long time. After the switch is opened, identical dielectric slabs are inserted in between the plates of both the capacitors and it is found that energy stored in capacitor B becomes nine times of the energy stored in capacitor A. What is the dielectric constant of the slab?

Answer

3

Explanation

Before the switch is opened, charge stored in both the capacitors is Q = CV. When it is opened then capacitor A becomes isolated from the battery so the amount of charge on it will remain same even after the slab is inserted in it. Whereas, in case of capacitor B, the battery remains connected and thus its potential difference remains constant. Now we can write the energy stored in both the capacitors after the switch is opened as follows:
`U_(B)=(1)/(2) (KC)V^(2) and U_(A)=((CV)^(2))/(2(KC))=(1)/(2) (CV^(2))/(K)`
Dividing the above equations we get
`(U_(B))/(U_(A))=((1)/(2)(KC)V^(2))/((CV^(2))/(2K))=K^(2)`
Given that ratio of energy stored in capacitors B and A is 9 : 1. So, we have
`(U_(B))/(U_(A))=9`
`rArr K^(2)=9`
Or `K=3`