A) 5
B) 10
C) \[6\sqrt{2}\]
D) \[6/\sqrt{2}\]
Correct Answer: C
Solution :
From Keplar third's law \[{{T}^{2}}\propto {{r}^{3}}\] Hence, \[T_{1}^{2}\propto r_{1}^{3}\] and \[T_{2}^{2}\propto r_{2}^{3}\] So, \[\frac{T_{2}^{2}}{T_{1}^{2}}=\frac{r_{2}^{3}}{r_{1}^{3}}\] \[=\frac{{{(3R)}^{3}}}{{{(6R)}^{3}}}\] or \[\frac{T_{2}^{2}}{T_{1}^{2}}=\frac{1}{8}\] \[T_{2}^{2}=\frac{1}{8}T_{1}^{2}\] \[T_{2}^{{}}=\frac{24}{2\sqrt{2}}=6\sqrt{2}h\]
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