A) 8 :1
B) 1 : 4
C) 4 : 1
D) 1: 8
Correct Answer: D
Solution :
From Keplars law \[{{T}^{2}}\propto {{R}^{3}}\] \[\therefore \] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{R_{1}^{3}}{R_{2}^{3}},\] Here, \[{{R}_{1}}=R\] and \[{{R}_{2}}=4R\] \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{R}{4R} \right)}^{3}}=\frac{1}{{{4}^{3}}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{{{4}^{3/2}}}=\frac{1}{{{2}^{3}}}=\frac{1}{8}\] \[{{T}_{1}}:{{T}_{2}}=1:8\]
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