Question

An automobile spring extends 0.2 m for 5000N 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 the $10\mu F$ capacitor at a potential difference of 10000V will be:

• A. 10000:1
• B. 1:1
• C. 1:2
• D. 2:1

Answer 1

The correct option is A 10000:1

Convert micro farads to farads :

$0.000001\times 10=0.00001$ farads

$P.E=0.5cV$

$P.E=0.5\times 0.00001\times 10000=0.1J$

P.E in the spring is :

$\frac{Force\times heightmoved}{5000\times 0.2=1000Joules}$

ratio of potential energy stored on the spring to P.E stored in the capacitor

$\frac{1000}{0.1}=\frac{1000}{1}$

$10000:1\to$ Required answer