A) Zero
B) \[\text{-}\frac{\text{3}}{\text{2}}\frac{\text{k}}{{{\text{a}}^{\text{2}}}}\]
C) \[\text{-}\frac{\text{k}}{\text{4 }{{\text{a}}^{\text{2}}}}\]
D) \[\frac{k}{2{{a}^{2}}}\]
Correct Answer: A
Solution :
\[U=\frac{-k}{2{{r}^{2}}}\Rightarrow F=\frac{-du}{dr}=\frac{k}{{{r}^{3}}}\] This force is providing centripetal acceleration. \[\frac{m{{v}^{2}}}{a}=\frac{k}{{{a}^{3}}}\] \[(\because r=a)\] \[\Rightarrow m{{v}^{2}}=k/{{a}^{2}}\] \[\Rightarrow \frac{1}{2}m{{v}^{2}}=\frac{k}{2{{a}^{2}}}\] \[\Rightarrow T.E.=\frac{-k}{2{{a}^{2}}}+\frac{k}{2{{a}^{2}}}=0\]
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