If `M` grams of ice is floating in a liquid of density `sigma_(L)`, then for its equilibrium, weight of the ice `=` thrust
or `Mg=V_(D)sigma_(L)g`
So the volume of the liquid displaced by the floating ice is
`V_(D)=(M//sigma_(L))`………i
Now if `M` grams of ice melts completely, water formed will have mass `M` grams (as mas is conserved). Now if `sigma_(w)` is the density of water the volume of water formed will be
`V_(F)=(M//sigma_(W))`............ii
Here the liquid is water i.e., `sigma_(L)=sigma_(W)`, so water displaced by floating ice is equal to the water formed by melting of the whole ice, and hence the level of water will remain unchanged. The following two points may be noted:
a. If `sigma_(L)gtsigma_(W)`, then `M//sigma_(L)gtM//sigma_(W)`, i.e., `V_(D)gtV_(F)`
i.e., water displaced by the floating ice will be lesser than water formed and so the level of the liquid in the beaker will rise.
b. If `sigma_(L)ltsigma_(W), M//sigma_(L)gtM//sigma_(W),` i.e., `V_(D)gtV_(F)`
i.e., water displaced by the floating ice will be more than water formed and so the liquid in the beaker will fall.