A) 1/2 yr
B) \[2\sqrt{2}yr\]
C) 4 yr
D) 8 yr
Correct Answer: B
Solution :
According to Keplars law \[{{T}^{2}}\propto {{R}^{3}}\] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] \[{{\left( \frac{1}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{R}{2R} \right)}^{3}}\] \[\frac{1}{T_{2}^{2}}=\frac{1}{8}\] \[{{T}_{2}}=2\sqrt{2}yr\]
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