A) \[\sqrt{r}\]
B) \[{{r}^{3/2}}\]
C) \[r\]
D) none of these
Correct Answer: A
Solution :
According to Kepler's third law \[{{T}^{2}}\propto {{r}^{3}}\] \[\Rightarrow \] \[T\propto {{r}^{3/2}}\] \[\omega =\frac{2\pi }{T}\] \[\therefore \] \[\omega \propto {{r}^{-3/2}}\] Now \[L=m{{r}^{2}}\omega \] \[\Rightarrow \] \[L\propto {{r}^{2}}\times {{r}^{-3/2}}\] \[\therefore \] \[L\propto {{r}^{1/2}}\]
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