Question

1. (a) a student places a hollow plastic ball and a solid plastic ball of the same size into water. the hollow ball floats, and the solid ball sinks. explain the difference in their behaviour using archimedes’ principle and the concept of density.

(b) water ice has a density of 0.91 g/cm³, so it will float in liquid water. imagine you have a cube of ice, 10 cm on a side.
(i) what is the cube’s weight?
(ii) what volume of liquid water must be displaced to support the floating cube?
(iii)how much of the cube is under the surface of the water?

Explanation:

Part (a)

Archimedes’ principle states that the buoyant force (\(F_b\)) on an object in a fluid is equal to the weight of the fluid displaced (\(F_b = W_{displaced}\)). Density (\(
ho\)) is mass (\(m\)) per unit volume (\(V\)), \(
ho=\frac{m}{V}\), and weight (\(W\)) is \(W = mg\) (where \(g\) is gravitational acceleration).

For the two balls (same size, so same volume \(V\) displaced initially):

• Hollow ball: Its average density (\(

ho_{hollow}\)) is less than water’s density (\(
ho_{water}\)). The weight of the hollow ball (\(W_{hollow}=
ho_{hollow}Vg\)) is less than the buoyant force (\(F_b=
ho_{water}Vg\)) when fully submerged, so it floats (displaces less water until \(F_b = W_{hollow}\)).

• Solid ball: Its density (\(

ho_{solid}\)) is greater than \(
ho_{water}\). Its weight (\(W_{solid}=
ho_{solid}Vg\)) is greater than \(F_b\) (even when fully submerged), so it sinks.

Part (b)

(i) Weight of the ice cube

Step 1: Calculate volume of ice cube

The ice cube is a cube with side \(s = 10\ \text{cm}\). Volume of a cube \(V = s^3\).
\(V = (10\ \text{cm})^3 = 1000\ \text{cm}^3\)

Step 2: Calculate mass of ice

Density of ice \(
ho_{ice}=0.91\ \text{g/cm}^3\). Using \(
ho=\frac{m}{V}\), rearrange to \(m=
ho V\).
\(m = 0.91\ \text{g/cm}^3\times1000\ \text{cm}^3 = 910\ \text{g}=0.91\ \text{kg}\) (since \(1\ \text{kg}=1000\ \text{g}\))

Step 3: Calculate weight of ice

Weight \(W = mg\), where \(g = 9.8\ \text{m/s}^2\) (or \(9.8\ \text{N/kg}\)).
\(W = 0.91\ \text{kg}\times9.8\ \text{N/kg}=8.918\ \text{N}\approx8.92\ \text{N}\)

(ii) Volume of water displaced

Step 1: Apply Archimedes’ principle

For a floating object, buoyant force equals weight (\(F_b = W_{ice}\)). The buoyant force is also \(F_b=
ho_{water}V_{displaced}g\), where \(
ho_{water}=1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3\).

Step 2: Solve for \(V_{displaced}\)

Set \(
ho_{water}V_{displaced}g = W_{ice}\). We know \(W_{ice}=mg=
ho_{ice}V_{ice}g\), so:
\(
ho_{water}V_{displaced}g=
ho_{ice}V_{ice}g\)
Cancel \(g\): \(
ho_{water}V_{displaced}=
ho_{ice}V_{ice}\)
\(V_{displaced}=\frac{
ho_{ice}}{
ho_{water}}V_{ice}\)

Substitute values: \(
ho_{ice}=0.91\ \text{g/cm}^3\), \(
ho_{water}=1\ \text{g/cm}^3\), \(V_{ice}=1000\ \text{cm}^3\)
\(V_{displaced}=\frac{0.91}{1}\times1000\ \text{cm}^3 = 910\ \text{cm}^3\)

(iii) Depth of ice under water

Answer:

Step 1: Relate displaced volume to submerged height

The ice cube has a square base with area \(A = s^2=(10\ \text{cm})^2 = 100\ \text{cm}^2\). The volume of water displaced (\(V_{displaced}\)) equals the volume of the submerged part of the cube, which is \(V_{submerged}=A\times h\) (where \(h\) is the height under water).

Step 2: Solve for \(h\)

Rearrange \(V_{submerged}=A\times h\) to \(h=\frac{V_{submerged}}{A}\). We know \(V_{submerged}=V_{displaced}=910\ \text{cm}^3\), \(A = 100\ \text{cm}^2\).
\(h=\frac{910\ \text{cm}^3}{100\ \text{cm}^2}=9.1\ \text{cm}\)

Final Answers

(a) Explanation as above.
(b) (i) \(\boldsymbol{8.92\ \text{N}}\) (or \(910\ \text{g}\times g\) with \(g = 9.8\ \text{N/kg}\))
(ii) \(\boldsymbol{910\ \text{cm}^3}\)
(iii) \(\boldsymbol{9.1\ \text{cm}}\)