Question

A 0.190-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.60 cm . The level of the glycerine is well below the top of the cylinder. If the ice completely melts, by what distance does the height of liquid in the cylinder change?

Answer 1

Weight of the displaced liquid equals the weight of the cube (or the mass of the displaced liquid equals the mass of the cube).
mass = density*volume
m = ?*V

m=0.190 kg

(1.26 is the Density for Glycerine in kg/L)

0.19 kg = 1.26 kg/L * V
V = 0.19 kg/1.26 kg/L = 0.1508 L (= dm^3 = displaced liquid)
Since the initial volume vi of the glycerol without the cube stays the same after the cube is added, the new volume (up to the surface only) in the cylinder after addition of the cube is the initial volume + the immersed volume
i.e., V = Vi + 0.1508 dm^3.

Now let the cube melt: 0.19 kg ice give 0.19 dm^3 water.
The volume in the cylinder is now vi + 0.19 dm^3.

The volume difference before, and after the melting, is then 0.19 - 0.1508 = 0.0392 dm^3
now find the height of 0.0392 dm^3 in the cylinder:
V = pir^2*h
h = V/(pir^2) with r in dm

radius =0.36m
h = 0.0392/(pi*0.36^2)
h = 0.09633 dm =0.963 cm > the liquid level is 0.963 cm higher after the melting.

I hope help you !!