A) Zero
B) \[\text{-}\frac{\text{k}}{\text{4 }{{\text{a}}^{\text{2}}}}\]
C) \[\frac{k}{2{{a}^{2}}}\]
D)
Correct Answer: A
Solution :
| [a] \[U=\frac{-k}{2{{r}^{2}}}\Rightarrow F=\frac{-du}{dr}=\frac{k}{{{r}^{3}}}\] |
| 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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