JEE Main On 08 April 2018 Question 5
Question: A particle is moving in a circular path of radius a under the action of an attractive potential $ \text{U=-}\frac{k}{2{r^{2}}} $ . Its total energy is: [JEE Main Online 08-04-2018]
Options:
A) Zero
B) $ \text{-}\frac{3}{2}\frac{k}{{a^{2}}} $
C) $ \text{-}\frac{k}{\text{4 }{a^{2}}} $
D) $ \frac{k}{2a^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ U=\frac{-k}{2r^{2}}\Rightarrow F=\frac{-du}{dr}=\frac{k}{r^{3}} $ This force is providing centripetal acceleration. $ \frac{mv^{2}}{a}=\frac{k}{a^{3}} $ $ (\because r=a) $
$ \Rightarrow mv^{2}=k/a^{2} $
$ \Rightarrow \frac{1}{2}mv^{2}=\frac{k}{2a^{2}} $
$ \Rightarrow T.E.=\frac{-k}{2a^{2}}+\frac{k}{2a^{2}}=0 $
