Density Worked ExamplesIGCSE Physics 0625

(a) Density is mass per unit volume. Write it as an equation, in symbols or in words.

(b) Rearrange to make volume the subject before substituting.

Substitute the mass in kg and the density in kg/m³, so the volume comes out in m³.

Round to 2 significant figures. The first two significant figures are the $1$ and the $8$; the next digit is $3$, so the $8$ stays.

Density needs two measurements: the mass and the volume. The mass is measured on a balance.

The rock is irregular, so its volume cannot be calculated from length measurements with a rule. Instead it is found by displacement: lower the rock into water in a measuring cylinder and read the rise in the water level.

So the student needs a balance and a measuring cylinder. A gives the mass but no volume. B would only work for a regular shape such as a cuboid. D gives the volume but no mass.

The answer is wanted in kg/m³, so convert every measurement to kg and m first. $1\text{ cm} = 0.01\text{ m}$ and $1\text{ mm} = 0.001\text{ m}$.

Volume of a cuboid = length × width × thickness.

Density = mass ÷ volume.

Round to 3 significant figures, matching the data, and give the unit.

Mass. Place the ornament on a balance that has been zeroed (tared) with nothing on the pan, and record its mass $m$.

First volume reading. Partly fill a measuring cylinder with water — enough to cover the ornament completely — and record the volume $V_1$, reading the bottom of the meniscus with your eye level with it.

Second volume reading. Lower the ornament gently into the water (for example on a thread, so no water splashes out) until it is fully submerged. Record the new reading $V_2$.

Calculation. The rise in level is the volume of water displaced, which equals the volume of the ornament. Then divide mass by volume.

(a) Compare the density of the object with the density of the liquid. Concrete is more dense than the lake water, so the block sinks.

(b) The canoe is not solid concrete: it encloses a large volume of air. Its average density is its total mass divided by the total volume it occupies.

Now compare with the water. The average density of the canoe is less than the density of the lake water, so the canoe floats.

Calculate the density of each liquid with $\rho = m/V$.

A less dense liquid floats on a more dense one, so the least dense liquid is at the top and the most dense at the bottom.

D orders the liquids by mass (46, 63, 80 g), not density — Z has the largest mass but not the largest density. B is the correct order upside down. A puts the medium-density liquid on top.

(a) This is about using the balance. Either check that it reads zero (tare it) before anything is placed on it, or place the cylinder in the centre of the pan.

(b) Each small division is $1\text{ cm}^3$ (ten divisions between $50$ and $60$). Read the bottom of the meniscus: it is level with the sixth mark above $50$.

(c) The mass of the liquid is the difference between the two balance readings.

Divide by the volume. The mass is in g and the volume in cm³, so the unit is g/cm³.

Round to 2 significant figures.

(a) The mass of all eight nuts is the difference between the balance readings. Then share it between eight.

(b) Each small division is $2\text{ cm}^3$. The level rises from $40\text{ cm}^3$ to $48\text{ cm}^3$, and that rise is the volume of all eight nuts.

(c) Density = mass ÷ volume for one nut (or for all eight — the ratio is the same).

(d) One nut would raise the water by only about $1\text{ cm}^3$ — half of one scale division — so the reading would be very uncertain. Eight nuts give a rise large enough to read, and dividing by eight reduces the uncertainty in the volume of one nut.

A further improvement: use a narrower measuring cylinder with smaller scale divisions, or use even more nuts. Repeating the whole measurement and calculating the mean would also help.

(a) Read both inner faces of the blocks to the nearest millimetre, then subtract.

(b) Substitute into the given formula.

(c) Mass of the ball = reading with the ball − reading of the empty dish.

Density = mass ÷ volume.

(d) Compare the densities: the wood is less dense than water, so the ball floats.

(a) Each line is straight and passes through the origin, so mass ∝ volume and the gradient is $m/V$, the density. Use a large triangle: read each line at $V = 50\text{ cm}^3$.

(b) Only B is less dense than water.

(c) Go across from $81\text{ g}$ on the mass axis to line A, then down to the volume axis. Check with the equation.

(a) Work out the volume $96.5\text{ g}$ of pure gold would have, and compare it with the measured volume.

Equivalently, the bracelet's density is $96.5 \div 6.0 = 16.1\text{ g/cm}^3$, which is less than $19.3\text{ g/cm}^3$, so it is not pure gold.

(b) Let the mass of gold be $g$ grams, so the mass of silver is $(96.5 - g)$ grams. Each metal's volume is its mass ÷ its density, and the volumes add to $6.0\text{ cm}^3$.

Evaluate $96.5 \div 10.5 = 9.190$ and collect the terms in $g$.

Solve.

Check: silver $= 96.5 - 73.5 = 23.0\text{ g}$. Volumes: $73.5 \div 19.3 = 3.81\text{ cm}^3$ and $23.0 \div 10.5 = 2.19\text{ cm}^3$, which add to $6.0\text{ cm}^3$. ✓

(a) When $V = 0$ there is no liquid, so the total mass is just the beaker. That is the intercept on the mass axis.

(b) Each extra $1\text{ cm}^3$ of liquid adds the same mass, so the gradient of the line is the density. Use two points on the line that are far apart, such as $(0, 62)$ and $(60, 137)$.

(c) $136.9 \div 60$ includes the $62\text{ g}$ of the beaker in the mass, so it is not the mass of the liquid alone. The line does not pass through the origin, so $M/V$ is not constant. (The liquid alone: $(136.9 - 62) \div 60 = 1.25\text{ g/cm}^3$.)

(d) Compare the densities: the bead is less dense than the liquid, so it floats.