A hollow iron ball floats completely immersed in water at `10^(@)C` temperature. What will happen if the temperature of both of them is raised to `50^(@)C?`

Explanation

The iron ball will sink in water if the temperature of the ball and of water is raised to `50^(@)C` from `10^(@)C.`
Let `d_(1)` be the density of water and `V_(1)` be the volume of the iron ball at `10^(@)C`
So, the apparent loss in weight of the ball at `10^(@)C=V_(1)d_(1)g.` Again, if `d_(2)` is the density of water and `V_(2)` is the volume of the iron ball at `50^(@)C`, the apparent loss in weight of the ball at `50^(@)C=V_(2)d_(2)g`
Now, `V_(2)=V_(1)[1+gamma_(i)(50-10)]`
`" ["gamma_(i)=` coefficient of volume expansion of iron]
and `d_(2)=d_(1)[1-gamma_(w)(50-10)]`
`" ["gamma_(w=` coefficient of volume expansion of water]
`therefore " " V_(2)d_(2)=V_(1)d_(1)[1+40gamma_(i)][1-40gamma_(w)]`
`" "=V_(1)d_(1)[1-40(gamma_(w)-gamma_(i))]`(approx.)
Since `gamma_(w) gt gamma_(i), V_(2)d_(2)g lt V_(1)d_(1)g`. Therefore, the loss in weight of the ball at `50^(@)C` temperature is less. Hence its apparent weight increases. So the ball will sink in water.