A) \[12\text{ }h\]
B) \[\text{6 }h\]
C) \[3\text{ }h\]
D) \[4\text{ }days\]
Correct Answer: C
Solution :
From Kepler's third law \[{{T}^{2}}\propto {{R}^{3}}\] or \[T\propto {{R}^{3/2}}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3/2}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{4{{R}_{1}}}{{{R}_{1}}} \right)}^{3/2}}\] \[{{T}_{2}}=\frac{{{T}_{1}}}{8}=\frac{29}{8}=3h\]
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