Physics Elasticity Question 48
Question: If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be
[CPMT 1976]
Options:
A) V/25
B) 5V
C) V/5
D) 25V
Show Answer
Answer:
Correct Answer: D
Solution:
$ U=\frac{1}{2}\left( \frac{YA}{L} \right),l^{2} $ \ $ U\propto l^{2} $
$ \frac{U _{2}}{U _{1}}={{\left( \frac{l _{2}}{l _{1}} \right)}^{2}}={{\left( \frac{10}{2} \right)}^{2}}=25 $
Therefore $ U _{2}=25U _{1} $ i.e. potential energy of the spring will be 25 V
BETA
