JEE Main 17 March 2021 Shift 1 Solved Paper

Given, capacitance of capacitor, C=14pF
Potential difference, V=12V
Energy of capacitor, Ei=‌

1

2

CV2
=‌

1

2

×14×12×12=1008pJ
Now, the charging battery is disconnected and a porcelain plate with K=7 is inserted between the plates
∴‌‌Ef=‌

Ei

7

⇒f=‌

1008

7

pj⇒f=144pj
Mechanical energy with which the plate would oscillate back and forth between the plates will be =(1008−144)pJ=864pJ