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

The gravity on moon is about one sixth of that on the earth
i.e. `g_(m)=g/t`
on earth water and air both exert buoyant force on the ball, but on moon as there is air so weight of the ball is balanced only buouant force of the water. Thus
On earth `mg=V_(w)rho_(w)g+V_(a)rho_(a)g`
or `m=V_(w)rho_(w)+V_(a)rho_(a)`.......ii
on moon
`m(g/6)=V_(2)^(')+V_(a)rho_(a)`............ii
or `m=V_(m)^(')rho_(w)`
From eqn i and ii we have `V_(w)^(')=V_(w)+(V_(a)rho_(a))/(rho_(w))`
Clearly `V_(w)^(')gtV_(w)`. That is the volume of the ball mmersed in water on the moon will be greater than that on the earth. Hence ball will sink slightly more in water on the moon.