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
Pick an option to check your answer.
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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