IGCSE Physics | Practice Test — Answers & Worked Solutions
Write down the definition of density.
Density is the mass per unit volume of a substance.
Write down the equation for density. Name each symbol and state its unit.
- $\rho$ (rho) = density, unit: $\text{g/cm}^3$ or $\text{kg/m}^3$
- $m$ = mass, unit: $\text{g}$ or $\text{kg}$
- $V$ = volume, unit: $\text{cm}^3$ or $\text{m}^3$
Write down the density equation rearranged to find: (a) mass and (b) volume.
State the density of water in g/cm³.
Name the laboratory instrument used to measure the volume of a liquid.
Explain in one or two sentences what is meant by volume by displacement.
Volume by displacement means finding the volume of a solid by measuring how much liquid it pushes aside when fully submerged. The volume of liquid displaced equals the volume of the solid.
State the density condition for an object to float in a liquid.
State the density condition for an object to sink in a liquid.
When using the displacement method, why must the solid sink completely to the bottom of the measuring cylinder?
If the solid only partially submerges (floats), the volume of water displaced is less than the full volume of the solid. The measured volume would be too small, giving a density that is too large.
State the rule for deciding which of two immiscible liquids will form the top layer. Supplement
A rectangular metal block: length 10 cm, width 5 cm, height 2 cm. Mass = 200 g. Calculate the density.
- Formula: $$\rho = \frac{m}{V}$$
- Find volume: $$V = 10 \times 5 \times 2 = 100\ \text{cm}^3$$
- Given: $m = 200\ \text{g}$, $V = 100\ \text{cm}^3$
- Substitute: $$\rho = \frac{200}{100}$$
- Answer: $$\rho = 2\ \text{g/cm}^3$$
100 cm³ of liquid has mass 92 g. Calculate the density.
- Formula: $$\rho = \frac{m}{V}$$
- Given: $m = 92\ \text{g}$, $V = 100\ \text{cm}^3$
- Substitute: $$\rho = \frac{92}{100}$$
- Answer: $$\rho = 0.92\ \text{g/cm}^3$$
Iron: density 7.8 g/cm³, volume 10 cm³. Calculate the mass.
- Formula: $$m = \rho \times V$$
- Given: $\rho = 7.8\ \text{g/cm}^3$, $V = 10\ \text{cm}^3$
- Substitute: $$m = 7.8 \times 10$$
- Answer: $$m = 78\ \text{g}$$
Material: density 2.5 g/cm³, mass 100 g. Calculate the volume.
- Formula: $$V = \frac{m}{\rho}$$
- Given: $m = 100\ \text{g}$, $\rho = 2.5\ \text{g/cm}^3$
- Express 2.5 as a fraction, then apply KCF (Keep, Change, Flip): $$V = 100 \div \frac{5}{2} = \frac{100}{1} \times \frac{2}{5}$$
- Cross-cancel — top-left 100 and bottom-right 5 share a factor of 5: $$V = \frac{\cancelto{20}{100}}{1} \times \frac{2}{\cancelto{1}{5}} = \frac{20 \times 2}{1 \times 1}$$
- Answer: $$V = 40\ \text{cm}^3$$
Rock in measuring cylinder: water rises from 30 cm³ to 50 cm³. Rock mass = 96 g. Calculate the density.
- Formula: $$\rho = \frac{m}{V}$$
- Find volume of rock: $$V = 50 – 30 = 20\ \text{cm}^3$$
- Given: $m = 96\ \text{g}$, $V = 20\ \text{cm}^3$
- Substitute: $$\rho = \frac{96}{20}$$
- Simplify by dividing numerator and denominator by 4: $$\rho = \frac{96 \div 4}{20 \div 4} = \frac{24}{5}$$ Convert: $5 \times 4 = 20$, remainder $4$; $\frac{4}{5} = 0.8$; so $\frac{24}{5} = 4.8$
- Answer: $$\rho = 4.8\ \text{g/cm}^3$$
Empty cylinder = 50 g. After adding liquid: volume = 40 cm³, total mass = 86 g. (a) Mass of liquid. (b) Density. (c) Mass of 80 cm³ of same liquid.
- Formula: $$\rho = \frac{m}{V}$$
- Given: $m = 36\ \text{g}$, $V = 40\ \text{cm}^3$
- Substitute: $$\rho = \frac{36}{40}$$
- Simplify by dividing by 4: $$\rho = \frac{36 \div 4}{40 \div 4} = \frac{9}{10} = 0.9\ \text{g/cm}^3$$
- Formula: $$m = \rho \times V$$
- Given: $\rho = 0.9\ \text{g/cm}^3$, $V = 80\ \text{cm}^3$
- Substitute: $$m = 0.9 \times 80$$
- Decompose: $(1 – 0.1) \times 80 = 80 – 8 = 72$, so $m = 72\ \text{g}$
Objects A (0.7 g/cm³), B (1.3 g/cm³), C (1.0 g/cm³) placed in water (1.0 g/cm³). State and explain what happens to each.
Floats. Its density ($0.7\ \text{g/cm}^3$) is less than the density of water ($1.0\ \text{g/cm}^3$).
Sinks. Its density ($1.3\ \text{g/cm}^3$) is greater than the density of water ($1.0\ \text{g/cm}^3$).
Suspended (neither floats nor sinks). Its density equals the density of water, so there is no tendency to rise or fall.
Metal block: 8 cm × 5 cm × 2 cm, mass 400 g. (a) Volume. (b) Density. (c) Float or sink in water?
- Formula: $$\rho = \frac{m}{V}$$
- Given: $m = 400\ \text{g}$, $V = 80\ \text{cm}^3$
- $$\rho = \frac{400}{80} = 5\ \text{g/cm}^3$$
The density of the block ($5\ \text{g/cm}^3$) is greater than the density of water ($1.0\ \text{g/cm}^3$).
Stone: mass 0.2 kg, volume 80 cm³. Calculate density in g/cm³. Convert mass first.
- Formula: $$\rho = \frac{m}{V}$$
- Convert mass: $$m = 0.2\ \text{kg} \times 1000 = 200\ \text{g}$$
- Given: $m = 200\ \text{g}$, $V = 80\ \text{cm}^3$
- Substitute: $$\rho = \frac{200}{80}$$
- Simplify by dividing numerator and denominator by 40: $$\rho = \frac{200 \div 40}{80 \div 40} = \frac{5}{2} = 2.5\ \text{g/cm}^3$$
- Answer: $$\rho = 2.5\ \text{g/cm}^3$$
Wooden ball: 0.65 g/cm³. Glass marble: 2.5 g/cm³. Both placed in water (1.0 g/cm³). Explain what happens to each.
Density of wooden ball ($0.65\ \text{g/cm}^3$) is less than density of water ($1.0\ \text{g/cm}^3$).
Density of glass marble ($2.5\ \text{g/cm}^3$) is greater than density of water ($1.0\ \text{g/cm}^3$).
Student finds density of a stone using displacement. (a) Describe method. (b) Water: 25 → 55 cm³. (c) Mass = 90 g, find density. (d) Float or sink?
- Use a balance to measure the mass of the stone.
- Fill a measuring cylinder with water and record the initial volume $V_1$.
- Carefully lower the stone into the water, making sure it is fully submerged. Record the new volume $V_2$.
- Volume of stone = $V_2 – V_1$.
- Calculate density using $\rho = \dfrac{m}{V}$.
- Formula: $$\rho = \frac{m}{V}$$
- Given: $m = 90\ \text{g}$, $V = 30\ \text{cm}^3$
- $$\rho = \frac{90}{30} = 3\ \text{g/cm}^3$$
Density of stone ($3\ \text{g/cm}^3$) > density of water ($1.0\ \text{g/cm}^3$).
Liquid density = 1.2 g/cm³. Sample mass = 120 g. (a) Find volume. (b) Find mass of 300 cm³. (c) Explain why density need not be re-measured.
- Formula: $$V = \frac{m}{\rho}$$
- Given: $m = 120\ \text{g}$, $\rho = 1.2\ \text{g/cm}^3$
- Express 1.2 as a fraction, then apply KCF: $$V = 120 \div \frac{6}{5} = \frac{120}{1} \times \frac{5}{6}$$
- Cross-cancel — top-left 120 and bottom-right 6 share a factor of 6: $$V = \frac{\cancelto{20}{120}}{1} \times \frac{5}{\cancelto{1}{6}} = \frac{20 \times 5}{1 \times 1}$$
- Answer: $$V = 100\ \text{cm}^3$$
- Formula: $$m = \rho \times V$$
- Given: $\rho = 1.2\ \text{g/cm}^3$, $V = 300\ \text{cm}^3$
- $$m = 1.2 \times 300$$
- Decompose: $(1 + 0.2) \times 300 = 300 + 60 = 360$
- Answer: $$m = 360\ \text{g}$$
Density is a fixed property of a pure substance. It does not change with the amount of liquid. The same liquid always has the same density ($1.2\ \text{g/cm}^3$), so we can use this value to calculate the mass of any volume of that liquid.
Objects in water (1.0 g/cm³): Cork 0.24, Ice 0.92, Pine 0.50, Rubber 1.2, Aluminium 2.7 (all g/cm³). (a) Which float? (b) Which sink? (c) Why does ice float on water?
Although ice and liquid water are the same substance (H&sub2;O), ice has a lower density ($0.92\ \text{g/cm}^3$) than liquid water ($1.0\ \text{g/cm}^3$). Because the density of ice is less than the density of liquid water, ice floats.
Supplement Three immiscible liquids: X = 1.6, Y = 0.7, Z = 1.1 (g/cm³). (a) Top layer? (b) Bottom layer? (c) Evaluate student’s claim about adding Liquid Z.
Liquid Y has the lowest density ($0.7\ \text{g/cm}^3$).
Liquid X has the highest density ($1.6\ \text{g/cm}^3$).
The student is incorrect. The position of each liquid depends only on its density, not on the amount (volume or mass) of liquid present. Liquid Z has density $1.1\ \text{g/cm}^3$, which is always less than Liquid X ($1.6\ \text{g/cm}^3$). Adding more Liquid Z increases its total mass and volume, but its density stays the same. Liquid X will always remain at the bottom regardless of how much Liquid Z is added.
Unknown metal: displacement method. (a) Describe steps. (b) Mass 237 g, water 50 → 80 cm³: find volume and density. (c) Float or sink? (d) Source of error.
- Use a balance to measure the mass of the metal sample. Record the mass.
- Fill a measuring cylinder with water and record the initial water level $V_1$. Read at the bottom of the meniscus, with your eye level with the surface.
- Carefully lower the metal sample into the water, making sure it is fully submerged.
- Record the new water level $V_2$.
- Calculate the volume of the metal: $V = V_2 – V_1$. Then calculate $\rho = m \div V$.
- Formula: $$\rho = \frac{m}{V}$$
- Given: $m = 237\ \text{g}$, $V = 30\ \text{cm}^3$
- Substitute: $$\rho = \frac{237}{30}$$
- Simplify: 237 and 30 are both divisible by 3.
Find $237 \div 3$: $3 \times 70 = 210$; $237 – 210 = 27$; $27 \div 3 = 9$; so $237 \div 3 = 79$ $$\rho = \frac{237 \div 3}{30 \div 3} = \frac{79}{10} = 7.9\ \text{g/cm}^3$$ - Answer: $$\rho = 7.9\ \text{g/cm}^3$$
Density of metal ($7.9\ \text{g/cm}^3$) > density of water ($1.0\ \text{g/cm}^3$).
Parallax error — if the eye is not level with the water surface when reading the volume, the reading will be incorrect. For example, if the eye is above the meniscus, the volume appears higher than the true value. This would make the calculated volume of the metal appear larger, giving a density that is lower than the actual value.
