Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of densities of earth $$({{\rho }_{e}})$$ and moon $$({{\rho }_{m}})$$ is $$\left( \frac{{{\rho }_{e}}}{{{\rho }_{m}}} \right)=\frac{5}{3}$$ then radius of moon Rm in terms of Re will be [MP PMT 2003]

Question

Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth.  If the ratio of densities of earth $$({{\rho }_{e}})$$ and moon $$({{\rho }_{m}})$$ is $$\left( \frac{{{\rho }_{e}}}{{{\rho }_{m}}} \right)=\frac{5}{3}$$ then radius of moon Rm in terms of Re will be    [MP PMT 2003]

TuteeHUB · Accepted Answer

$$\frac{1}{6}{{R}_{e}}$$

TuteeHUB · Answer

$$\frac{5}{18}{{R}_{e}}$$

TuteeHUB · Answer

$$\frac{1}{6}{{R}_{e}}$$

TuteeHUB · Answer

$$\frac{3}{18}{{R}_{e}}$$

TuteeHUB · Answer

$$\frac{1}{2\sqrt{3}}{{R}_{e}}$$

1.	Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of densities of earth \[({{\rho }_{e}})\] and moon \[({{\rho }_{m}})\] is \[\left( \frac{{{\rho }_{e}}}{{{\rho }_{m}}} \right)=\frac{5}{3}\] then radius of moon Rm in terms of Re will be [MP PMT 2003]
A.	\[\frac{5}{18}{{R}_{e}}\]
B.	\[\frac{1}{6}{{R}_{e}}\]
C.	\[\frac{3}{18}{{R}_{e}}\]
D.	\[\frac{1}{2\sqrt{3}}{{R}_{e}}\]
Answer» B. \[\frac{1}{6}{{R}_{e}}\]

Discussion

No Comment Found

Related MCQs

Reply to Comment