Question

A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 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 gylcerine before the ice melted?

Answer 1

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

$mg = \rho V_{displaced} g$

$V_{displaced} = \frac{m}{\rho}$

$V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3$

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

$V\rho_w = m_{ice}$

$V = \frac{0.200}{1000} = 2\times 10^{-4} m^3$

So here extra volume that rise in the level will be given as

$\Dleta V = V - V_{displaced}$

$\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}$

$(\pi (0.039^2) h = 0.41 \times 10^{-4}$

$h = 0.86 cm$

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice