An automobile spring extends 0.2 m for 5000 N load.The ratio of potential energy stored in this spring when it has been compressed by 0.2 m to the potential energy stored in 10 microF capacitor at a potential difference of 10000 v Will be

Question

1-1/4
2-1
3-1/2
4-2

Answer 1

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Answer 2

First of all

Answer 3

For spring F=1/2kx2 and potential energy U=1/2Fx=0.5*0.2*5000=500J For capacitor potential energy U'=1/2CV2=0.5*10*[10]-6*10000*10000=500J. Therefore ratio of potential energy is 1:1