A) 16
B) 4
C) \[4\sqrt{27}\]
D) \[4\sqrt{8}\]
Correct Answer: C
Solution :
According to Keplers third law \[{{T}^{2}}\propto {{R}^{3}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3/2}}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{3R}{R} \right)}^{3/2}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{27}\] \[\therefore \] \[{{T}_{2}}=\sqrt{27}{{T}_{1}}=\sqrt{27}\times 4\] \[=4\sqrt{27}h\]
You need to login to perform this action.
You will be redirected in
3 sec
