A) \[D{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{1/2}}days\]
B) \[D{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{3/2}}days\]
C) \[D{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2/3}}days\]
D) \[D\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)days\]
Correct Answer: B
Solution :
According to Kepler's third law, \[{{T}^{2}}\propto {{r}^{3}}\] or \[T\propto {{r}^{3/7}}\] \[\therefore \frac{D'}{D}={{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{3/2}}\] or \[D'=D{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{3/2}}days\]
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