Answer :
The question is incomplete. Here is the complete question.
A 0.150 kg cube fo ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.10 cm. The level of the glycerine is well below the top of the cylinder.
Part A
If the ice completely melts, by what distance does the height of liquid in the cylinder change? Express your answer with the appropriate units.
Part B
Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the glycerine before the ice melted?
Answer: Part A: [tex]\Delta h=[/tex] 327 mm
Part B: Rise
Explanation: When an object is submerged on a fluid, the fluid exerts a upward force on the object called Buoyant Force. It can be calculated by the following relation:
[tex]B=\rho_{fluid}gV[/tex]
Part A
Because the ice cube is floating on glycerin, its forces are on equilibrium, which means buoyant and weight are equal:
[tex]\rho_{fluid}gV=mg[/tex]
Density of glycerine is 1260 kg/m³.
Volume of a cylinder is [tex]V=\pi.r^{2}.h[/tex]
Solving
[tex]\rho_{fluid}gV=mg[/tex]
[tex]\rho_{g}.V=m[/tex]
[tex]\rho_{g}.\pi.r^{2}.h_{1}=m[/tex]
[tex]h_{1}=\frac{m}{\rho_{g}.\pi.r^{2}}[/tex]
[tex]h_{1}=\frac{0.15}{1260.\pi.(0.055)^{2}}[/tex]
[tex]h_{1}=1.25[/tex] m
After melted, the volume of a mass of water created is
[tex]\rho_{w}=\frac{m}{V}[/tex]
[tex]V=\frac{m}{\rho_{w}}[/tex]
Density of water is 10³ kg/m³.
[tex]\pi.r^{2}.h_{2}=\frac{m}{\rho_{w}}[/tex]
[tex]h_{2}=\frac{m}{\rho_{w}.\pi.r^{2}}[/tex]
[tex]h_{2}=\frac{0.15}{1000.(0.055)^{2}.\pi}[/tex]
[tex]h_{2}=[/tex] 1.58 m
The change in heights is
[tex]\Delta h=h_{2}-h_{1}[/tex]
[tex]\Delta h=[/tex] 0.327 m
The distance by which the height of liquid in the cylinder change is 327mm.
Part B:
Water has more density than ice. So, when melted, water will take the place of glycerin more than ice and so, liquid will rise on the cylinder.