question_answer

A particle of mass m revolves round a horizontal circle of radius r under the influence of a K centripetal force equal to \[-\frac{k}{2r}\] , where K stands for a constant. What is the total energy possesses by this particle?

A)  \[+\frac{k}{r}\]                                

B)  \[-\frac{k}{2r}\]  

C) \[-\frac{\text{ 2K}}{r}\]                

D)         \[+\frac{\text{ K}}{{{r}^{2}}}\]

Correct Answer: B

Solution :

 Potential energy,\[U=-\delta F.dr=-\delta \frac{K}{{{r}^{2}}}dr.=-\frac{K}{r}\] Kinetic energy\[,\]\[K=\frac{1}{2}\].                 \[m{{v}^{2}}=\frac{K}{2r}\left( \operatorname{since}\frac{m{{v}^{2}}}{r}=\frac{K}{{{r}^{2}}}\Rightarrow {{v}^{2}}=\frac{K}{mr} \right)\] \[\therefore \]Total energy\[=-\frac{K}{r}+\frac{K}{2r}=-\frac{K}{2r}\]