Q. 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
Explanation
(b) When a force of $F$ Newton is applied the potential energy is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U = \frac{1}{2}Fx$
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Energy stored by capacitor is $\frac{1}{2}CV^{2}$
$\therefore\ $Ratio is $\frac{\frac{1}{2}Fx}{\frac{1}{2}CV^{2}} = \ \frac{5000 \times 0.2}{10 \times 10^{- 6} \times \left( 10^{4} \right)^{2}} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.