A) \[2mk{{r}^{2}}t\]
B) \[mk{{r}^{2}}{{t}^{2}}\]
C) \[m{{k}^{2}}{{r}^{2}}t\]
D) \[m{{k}^{2}}r{{t}^{2}}\]
Correct Answer: C
Solution :
\[{{a}_{c}}=\frac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}}\] \[KE=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{k}^{2}}{{r}^{2}}{{t}^{2}}\] According to work-energy theorem, change in kinetic energy is equal to work done. \[\therefore \] \[W=\frac{1}{2}m{{k}^{2}}{{r}^{2}}{{t}^{2}}\] Thus, power delivered to the particle, \[P=\frac{dW}{dt}=m{{k}^{2}}{{r}^{2}}t\]You need to login to perform this action.
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