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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2018
  • Medium
  • MCQ

A particle is moving in a circular path of radius a under the action of an attractive potential \(U = - \frac{k}{{2{r^2}}}\) Its total energy is

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Answer & Solution

Correct Answer

(B) Zero

Step-by-step Solution

Detailed explanation

\(F = \frac{{\partial u}}{{\partial r}}\,\hat r = \frac{K}{{{r^3}}}\,\hat r\)

Since particle is moving in circular path

\(\begin{gathered}
  F = \frac{{m{v^2}}}{r} = \frac{K}{{{r^3}}} \Rightarrow m{v^2} = \frac{K}{{{r^2}}} \hfill \\
  \therefore \,K.E. = \frac{1}{2}m{v^2} = \frac{K}{{2{r^2}}} \hfill \\ 
\end{gathered} \)

Total enerygy \(= P.E. + K.E.\)

\( =  - \frac{K}{{2{r^2}}} + \frac{K}{{2{r^2}}} = zero\,\,\left( {\because \,P.E. = \frac{K}{{2{r^2}}}given} \right)\)

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