A) 16
B) 4
C) \[4\sqrt{27}\]
D) 4\[\sqrt{8}\]
Correct Answer: C
Solution :
According to Kepler's third law \[{{T}_{2}}\propto {{R}^{3}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3/2}}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{3R}{R} \right)}^{3/2}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{27}\] \[\therefore \] \[{{T}_{2}}=\sqrt{27}{{T}_{1}}=\sqrt{27}\times 4=4\sqrt{27}h\]
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