A) \[2\pi m{{k}^{2}}{{r}^{2}}t\]
B) \[m{{k}^{2}}{{r}^{2}}t\]
C) \[\frac{m{{k}^{4}}{{r}^{2}}{{t}^{5}}}{3}\]
D) Zero
Correct Answer: B
Solution :
Here the tangential acceleration also exits which requires power. Given that \[{{a}_{C}}={{k}^{2}}r{{t}^{2}}\] and \[{{a}_{C}}=\frac{{{v}^{2}}}{r}\]\ \[\frac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}}\] or \[{{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}}\] or \[v=krt\] Tangential acceleration \[a=\frac{dv}{dt}=kr\] Now force \[F=m\times a=mkr\] So power \[P=F\times v=mkr\times krt=m{{k}^{2}}{{r}^{2}}t\]You need to login to perform this action.
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