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 `((rho_(e))/(rho_(m)))=5/3` then radius of moon `(R_(m))` in terms of `R_(e)` will be

Options

A. `5/18 R_(e)`
B. `1/6 R_(e)`
C. `3/18 R_(e)`
D. `1/(2sqrt(3)) R_(e)`

Answer

A

Explanation

`(g_(m))/(g_(e))=1/6, implies (rho_(m)R_(m))/(rho_(e)R_(e))=1/6`
`3/5.(R_(m))/(R_(e))=1/6 implies R_(m)=5/18 R_(e)`