These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This is the algebra-based course, so no calculus is needed. Use g = 9.8 m/s², the density of fresh water 1000 kg/m³ and the density of air 1.2 kg/m³. Round final answers to 2 significant figures unless told otherwise. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
A cooking oil has a density of 0.85 g/cm³. What is its density in SI units?
- (A) 8.5 × 10⁻⁴ kg/m³
- (B) 85 kg/m³
- (C) 850 kg/m³
- (D) 8.5 × 10⁵ kg/m³
Answer and explanation
Answer: (C). 1 g/cm³ = 10⁻³ kg ÷ 10⁻⁶ m³ = 1000 kg/m³, so 0.85 g/cm³ = 0.85 × 1000 = 850 kg/m³. This is a little less than water, which fits an oil.
- (A) divides by 1000 instead of multiplying: the conversion was done the wrong way round.
- (B) uses a factor of 100 (from 1 m = 100 cm) without cubing it and without converting grams to kilograms.
- (D) converts cm³ to m³ (× 10⁶) but forgets to convert grams to kilograms (÷ 1000).
Question 2 (multiple choice · foundation)
A student pours 200 mL of water from a tall, narrow glass into a wide, shallow dish. None is spilled. Which of the water’s properties change?
- (A) Its shape only
- (B) Its shape and its volume
- (C) Its volume and its density
- (D) Its shape, its volume and its density
Answer and explanation
Answer: (A). Water is a liquid. Its particles are close together and can slide past each other, so it has no fixed shape and takes the shape of the dish. But the particles are already touching, so the volume stays 200 mL. Mass and volume are both unchanged, so the density ρ = m / V is unchanged too.
- (B) treats the liquid like a gas, which spreads out to fill its container.
- (C) and (D) confuse a change of shape with a change of density. Spreading the water out does not change how much mass is in each cubic centimetre.
Question 3 (multiple choice · core)
Cube X has side length L and mass m. Cube Y has side length 2L and mass 4m. Each cube is made of one uniform material. What is the ratio ρ_Y / ρ_X of their densities?
- (A) 1/2
- (B) 2
- (C) 4
- (D) 1/8
Answer and explanation
Answer: (A). Volume of a cube = side³, so V_Y = (2L)³ = 8L³ = 8V_X. Then ρ_Y / ρ_X = (4m / 8V_X) ÷ (m / V_X) = 4 / 8 = 1/2. Cube Y has four times the mass, but eight times the volume.
- (B) divides the mass ratio (4) by the side-length ratio (2), as if volume grew in proportion to length instead of length cubed.
- (C) compares masses only. That would be true only if the volumes were equal.
- (D) compares volumes only and forgets the mass also changed.
Question 4 (multiple choice · core)
Which statement best explains why a sample of gas can be compressed much more easily than the same mass of liquid?
- (A) Gas particles are much smaller than liquid particles.
- (B) Gas particles are far apart compared with their size and interact only when they collide, so most of a gas’s volume is empty space.
- (C) Gas particles have no mass, so nothing resists squeezing them.
- (D) Liquid particles do not move, so they cannot be pushed closer together.
Answer and explanation
Answer: (B). Compressing a gas pushes its particles into some of the empty space between them. Liquid particles already touch their neighbours, so pushing them closer is very hard.
- (A) is not the reason. Water vapour and liquid water contain the same molecules; only the spacing differs.
- (C) is false. Air has a density of about 1.2 kg/m³, so its particles do have mass.
- (D) is false. Liquid particles move and slide past each other; that is why liquids flow.
Question 5 (calculation · core)
A car’s fuel tank holds 45 L of petrol. The petrol has a density of 740 kg/m³.
(a) Calculate the mass of petrol in a full tank. (b) Calculate the weight of this petrol. (c) The same tank is instead filled with fresh water. By what factor is the mass of water larger than the mass of petrol?
Worked solution
(a) Volume: V = 45 L = 45 × 10⁻³ m³ = 0.045 m³. Mass: m = ρV = 740 kg/m³ × 0.045 m³ = 33.3 kg ≈ 33 kg.
(b) Weight = mg = 33.3 kg × 9.8 m/s² = 326 N ≈ 3.3 × 10² N.
(c) Mass of water = 1000 × 0.045 = 45 kg. Ratio = 45 ÷ 33.3 = 1.35 (or ρ_water / ρ_petrol = 1000 / 740). For equal volumes, mass is proportional to density.
Suggested mark points (3): 1 for converting litres to m³ and getting 33 kg; 1 for weight with unit N; 1 for the ratio 1.35 with the reason “same volume, so mass ∝ density”.
Common error: using V = 45 m³, which gives 33,300 kg — the mass of a loaded lorry, not a tank of fuel.
Question 6 (constructed response · stretch)
A student wants to find the density of a salt solution. She has a measuring cylinder, an electronic balance and a supply of the solution. She does not know the mass of the empty cylinder.
(a) Describe a procedure that uses a graph to find the density without needing to measure the empty cylinder separately. Say what to plot on each axis.
She records these results (total mass = cylinder + solution):
| Volume of solution, V (mL) | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|
| Total mass, M (g) | 112.6 | 135.5 | 157.4 | 180.6 | 202.9 |
(b) Plot the data (or work from a best-fit line) and find the density of the solution in kg/m³. (c) State what the vertical intercept of the line represents, and give its value. (d) A classmate divides 112.6 g by 20 mL and gets 5.6 g/mL. Explain why this is wrong.
Worked solution and suggested Marlbridge rubric
(a) Place the empty cylinder on the balance. Pour in about 20 mL of solution, read the volume at eye level and record the total mass. Add more solution in steps of about 20 mL, recording volume and total mass each time. Plot total mass M (vertical, g) against volume V (horizontal, mL). The cylinder’s mass is the same in every reading, so it shifts the line up but does not change its slope. The slope is the density.
(b) The points lie on a straight line. Using two points on the best-fit line, (20 mL, 112.7 g) and (100 mL, 202.9 g): slope = (202.9 − 112.7) g ÷ (100 − 20) mL = 90.2 ÷ 80 = 1.13 g/mL. Since 1 mL = 1 cm³, ρ = 1.13 g/cm³ = 1.1 × 10³ kg/m³ (1130 kg/m³ to 3 s.f.). A salt solution slightly denser than pure water is reasonable.
(c) The intercept is the total mass when V = 0: the mass of the empty cylinder, about 90 g.
(d) 112.6 g includes the cylinder’s 90 g, so it is not the mass of 20 mL of solution. Only the solution’s mass (about 22.6 g) should be divided by 20 mL, which gives about 1.13 g/mL.
| Point | What earns it |
|---|---|
| 1 | Measures total mass for several different volumes of solution |
| 1 | Plots total mass against volume (axes named) and states that slope = density |
| 1 | Density 1.1 × 10³ kg/m³ from the slope, with correct unit conversion |
| 1 | Intercept identified as the mass of the empty cylinder, about 90 g |
| 1 | Explains that the single-point ratio includes the cylinder’s mass |
Alternative method. Subtracting consecutive readings (about 22–23 g for each extra 20 mL) and averaging also gives 1.13 g/mL and earns the density point, but the graph is the method the question asks for in (a).
Question 7 (constructed response · stretch)
Two sealed syringes each start with 60 mL of fluid. Syringe P contains air (density 1.2 kg/m³). Syringe Q contains water. A student pushes each plunger as hard as she can. The air in P is squeezed to 30 mL. The water in Q stays at 60 mL as far as she can measure.
A classmate claims: “Density is a property of a material, so the density of the air must still be 1.2 kg/m³.”
(a) Calculate the mass of air in syringe P, and its density after it is squeezed. (b) Evaluate the classmate’s claim. Use the particle picture of a gas and of a liquid in your answer. (c) Which syringe’s contents behave like an ideal fluid? Give a reason.
Model answer and suggested Marlbridge rubric
(a) V = 60 mL = 60 × 10⁻⁶ m³. Mass = ρV = 1.2 × 60 × 10⁻⁶ = 7.2 × 10⁻⁵ kg (0.072 g). The syringe is sealed, so the mass stays the same. New density = 7.2 × 10⁻⁵ kg ÷ 30 × 10⁻⁶ m³ = 2.4 kg/m³: double the starting value.
(b) The claim is wrong for the air. Density is mass ÷ volume. The mass of air is fixed, but its volume halved, so its density doubled. In a gas, the particles are far apart with mostly empty space between them, so pushing the plunger moves them closer together. The claim is reasonable for the water: liquid particles already touch, so the volume (and so the density) barely changes even under a large push. The idea “density is a property of a material” holds well for liquids and solids, but a gas’s density depends on how much it is compressed.
(c) The water (Q). An ideal fluid is incompressible, and the water kept its volume and density. The air in P was compressed, so it does not fit the incompressible part of the model.
| Point | What earns it |
|---|---|
| 1 | Mass of air 7.2 × 10⁻⁵ kg, with unit conversion shown |
| 1 | New density 2.4 kg/m³, using constant mass |
| 1 | Explains with the particle picture why the gas compresses (spacing, empty space) |
| 1 | Explains why the liquid does not (particles already close) |
| 1 | Identifies water as the ideal fluid because it is incompressible |
A correct ratio argument (“same mass, half the volume, so twice the density”) without the number earns the second point.
How did you do?
- Q1 or Q5 wrong: re-read “Converting units” and Worked example 1 in the study guide.
- Q2 or Q4 wrong: revisit “Why solids, liquids and gases behave differently” and Figure 1.
- Q3 wrong: remember volume scales with the cube of length. See the misconception “A smaller piece has a smaller density”.
- Q6 incomplete: work through Worked example 2 and Figure 2 (slope of a mass–volume graph).
- Q7 incomplete: go back to “The ideal-fluid model”.
Then tick off the topic checklist.