question_answer

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 a 10 \[\mu F\]capacitor at a potential difference of 10000 V will be

A)  1/4           

B)  1

C)  1/2           

D)  2

Correct Answer: B

Solution :

When a force of F newton is applied the potential energy is given by                 \[U=\frac{1}{2}Fx\] Energy stored by capacitor is \[\frac{1}{2}C\,{{V}^{2}}\]. \[\therefore \] Ratio is \[\frac{\frac{1}{2}Fx}{\frac{1}{2}C\,{{V}^{2}}}=\frac{5000\times 0.2}{10\times {{10}^{-6}}\times {{({{10}^{4}})}^{2}}}=1\]