A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{6}\]
Correct Answer: C
Solution :
From Keplers law \[{{T}^{2}}\propto {{R}^{3}}\] \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3/2}}={{\left( \frac{R}{R/4} \right)}^{3/2}}\] \[={{(4)}^{3/2}}={{(2)}^{3}}=8\] \[\therefore \] \[{{T}_{2}}=\frac{{{T}_{1}}}{8}.\] Hence, the length of the day is reduced by\[\frac{1}{8}.\]
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