A ball floats on the surface of water in a container exposed to the atmosphere. Will the ball remain immersed at its initial depth or will it sink or rise somewhat if the container is shifted to the moon?

Explanation

We know that the gravity on moon is about `1//6th` of earth. But the gravity has no effect on the equilibrium of floating body as weight and upthrust both will change by same factor. We also know that moon has no atmosphere whil earth has atmosphere. Therefore, on earth, the weight of the floating ball is balanced by upthrust from water and air, i.e.,
`W=mg = V_(w)rho_(w)g +V_(a)rho_(a)g`
or, `m=V_(w)rho_(w)+V_(a)rho_(a)` ...(i)
While on moon, `m=V_(w) rho_(w)` ...(ii)
From (i) and (ii)
`V_(w)^(') = V_(w) +(V_(a)rho_(a))/(rho_(w)), i.e., V_(w)^(') gt V_(w)`
It means the volume of the ball in water at moon will be more than on earth i.e. the ball will sink a little more.