Density: Question 4

Worked solution

Part (a): Where does the sphere settle?

Compare the sphere’s density with each liquid in turn.

• Sphere vs liquid P: $1.05\text{ g/cm}^3 > 0.80\text{ g/cm}^3$, so the sphere is denser than liquid P and sinks through it.
• Sphere vs liquid Q: $1.05\text{ g/cm}^3 < 1.26\text{ g/cm}^3$, so the sphere is less dense than liquid Q and floats on it.

Putting these together, the sphere sinks through liquid P but cannot sink into liquid Q, so it comes to rest at the boundary between the two liquids, resting on top of liquid Q.

Part (b): Mass of the sphere

Rearrange the density formula to make mass the subject:

$\rho = \frac{m}{V} \quad\Rightarrow\quad m = \rho V$

Substitute the sphere’s density and volume:

$m = 1.05\text{ g/cm}^3 \times 12.0\text{ cm}^3$

$m = \boxed{12.6\text{ g}}$

Part (c): Converting the density of liquid Q

To convert $\text{g/cm}^3$ to $\text{kg/m}^3$, note that:

$1\text{ g} = 0.001\text{ kg}, \qquad 1\text{ cm}^3 = (0.01\text{ m})^3 = 10^{-6}\text{ m}^3$

$1\text{ g/cm}^3 = \frac{0.001\text{ kg}}{10^{-6}\text{ m}^3} = 1000\text{ kg/m}^3$

So every value in $\text{g/cm}^3$ is multiplied by $1000$ to convert to $\text{kg/m}^3$:

$1.26\text{ g/cm}^3 \times 1000 = \boxed{1260\text{ kg/m}^3}$

Final answers

• (a) The sphere rests at the interface between liquid P and liquid Q (sinks through P, floats on Q).
• (b) Mass $=$ $12.6\text{ g}$
• (c) Density of liquid Q $=$ $1260\text{ kg/m}^3$