A rubber ball floats on the surface of water contained in a beaker, exposed to atmosphere certain volume immersed inside the water. If the whole arrangement is shifted to the ball remain immersed at the same depth?

Explanation

Gravity on the surface of moon is 1/6th of that on the earth surface. But the effect of (a) weight of the ball (b) upthrust. So, the equilibrium of floating rubber ball will remain undisturbed.
On earth, weight of the rubber ball will be balanced by upthrust from water and air.
weight of the ball `=mg=V_(w) p_(w) +V_(a) p_(a) g`
`V_(w)="volume of water displaced"`
`p_(w)="density of water"`
`V_(a)="volume of air displaced"`
`p_(a)="density of air"`
`m=V_(w)p_(w)+V_(a)p_(a)`
Now, on the moon, there is not atomosphere, So,
`mg=V_(w)^(.) p_(w) g`
`m=V_(w)^(.) p_(w)`
`V_(w)^(.)="Volume of water displaced on moon"`
Comparing (i) and (ii), we get
`V_(w) p_(w)=V_(a)p_(a)=V_(w)^(.)p_(w)`
`rArr V_(w)^(.)=V_(w)+(V_(a) p_(a))/(p_(w))`
`rArr V_(w)^(.) gt V_(w)`
So, the volume of ball immersed inside the water is greater on moon. So, the ball will likely to get a little more inside the water when taken on moon.