A Level Physics 4.3 Density and Pressure MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 4.3 Density and Pressure – density, hydrostatic pressure and upthrust, at full AS Level standard. Every question is original, with a fully worked explanation. Take g = 9.8 m s-2 in all calculations.
Density
Q1. A metal block has mass 0.54 kg and volume 200 cm3. What is its density?
- (A) 2.7 kg m-3
- (B) 270 kg m-3
- (C) 2700 kg m-3
- (D) 27 000 kg m-3
Show Answer
Answer: C – Convert the volume first: 200 cm3 = 200 × 10-6 m3 = 2.0 × 10-4 m3. ρ = m / V = 0.54 / (2.0 × 10-4) = 2700 kg m-3 (the density of aluminium). Option A is the value in g cm-3 – the classic unit trap.
Q2. The density of an oil is 0.80 g cm-3. What is this density in kg m-3?
- (A) 800 kg m-3
- (B) 0.80 kg m-3
- (C) 80 kg m-3
- (D) 8000 kg m-3
Show Answer
Answer: A – 1 g cm-3 = 1000 kg m-3, because 1 g = 10-3 kg and 1 cm3 = 10-6 m3, so the conversion factor is 10-3 / 10-6 = 1000. Hence 0.80 × 1000 = 800 kg m-3.
Q3. An alloy is made by melting together 2.0 kg of metal X (density 8000 kg m-3) and 3.0 kg of metal Y (density 6000 kg m-3). Assuming the volumes add, what is the density of the alloy?
- (A) 7000 kg m-3
- (B) 6700 kg m-3
- (C) 6900 kg m-3
- (D) 14 000 kg m-3
Show Answer
Answer: B – Work with total mass and total volume. VX = 2.0 / 8000 = 2.5 × 10-4 m3; VY = 3.0 / 6000 = 5.0 × 10-4 m3. Total volume = 7.5 × 10-4 m3, total mass = 5.0 kg. ρ = 5.0 / (7.5 × 10-4) ≈ 6700 kg m-3. Option A (the simple average of the two densities) is wrong because the two metals contribute different volumes.
Q4. An irregular stone of mass 78 g is lowered into a measuring cylinder. The water level rises from 50 cm3 to 80 cm3. What is the density of the stone?
- (A) 2.6 kg m-3
- (B) 260 kg m-3
- (C) 26 000 kg m-3
- (D) 2600 kg m-3
Show Answer
Answer: D – The displaced volume equals the volume of the stone: 80 − 50 = 30 cm3. ρ = m / V = 78 / 30 = 2.6 g cm-3 = 2.6 × 1000 = 2600 kg m-3.
Q5. A uniform block is cut into two equal halves. How does the density of each half compare with the density of the original block?
- (A) It is the same, because mass and volume both halve in the same ratio
- (B) It is half the original density
- (C) It is twice the original density
- (D) It depends on which half is measured
Show Answer
Answer: A – Density is an intensive property: it does not depend on the size of the sample. Each half has half the mass and half the volume, so m / V is unchanged.
Q6. Which of the following objects will sink when placed in water of density 1000 kg m-3?
- (A) Cork, density 240 kg m-3
- (B) Ice, density 920 kg m-3
- (C) Wood, density 600 kg m-3
- (D) Brick, density 1200 kg m-3
Show Answer
Answer: D – An object sinks if its density is greater than the density of the liquid, because even when fully immersed the maximum upthrust (weight of water displaced) is less than its weight. Only the brick, at 1200 kg m-3, is denser than water; the others float.
Q7. A tank of volume 0.050 m3 is completely filled with a liquid of density 800 kg m-3. What is the mass of the liquid?
- (A) 16 kg
- (B) 40 kg
- (C) 400 kg
- (D) 0.0625 kg
Show Answer
Answer: B – Rearrange ρ = m / V to give m = ρV = 800 × 0.050 = 40 kg. Option D comes from dividing instead of multiplying.
Q8. Samples of the same metal are measured: 10 cm3 has mass 27 g, 20 cm3 has mass 54 g and 40 cm3 has mass 108 g. What is the density of the metal?
- (A) 270 kg m-3
- (B) 2700 kg m-3
- (C) 27 kg m-3
- (D) 0.37 kg m-3
Show Answer
Answer: B – Every sample gives the same ratio: 27 / 10 = 2.7 g cm-3 (the gradient of a mass-volume graph). Converting, 2.7 × 1000 = 2700 kg m-3. Option D is the reciprocal, V / m.
Q9. 100 cm3 of water (density 1000 kg m-3) is mixed with 100 cm3 of oil (density 800 kg m-3). The total volume is 200 cm3. What is the average density of the mixture?
- (A) 800 kg m-3
- (B) 950 kg m-3
- (C) 900 kg m-3
- (D) 1800 kg m-3
Show Answer
Answer: C – Mass of water = 1000 × 100 × 10-6 = 0.100 kg; mass of oil = 800 × 100 × 10-6 = 0.080 kg. Total mass = 0.180 kg in 200 × 10-6 m3, so ρ = 0.180 / (2.0 × 10-4) = 900 kg m-3. Here the simple average works only because the two volumes are equal.
Q10. A block has density 950 kg m-3. In which liquid will the block float?
- (A) Liquid of density 850 kg m-3
- (B) Liquid of density 900 kg m-3
- (C) Liquid of density 940 kg m-3
- (D) Liquid of density 1200 kg m-3
Show Answer
Answer: D – Floating requires the liquid to be denser than the block, so that the weight of liquid displaced can equal the block’s weight before the block is fully submerged. Only the liquid of density 1200 kg m-3 is denser than 950 kg m-3; in the others the block sinks.
Pressure in Fluids
Q11. A force of 600 N acts on a flat surface of area 0.020 m2. What is the pressure on the surface?
- (A) 3.0 × 104 Pa
- (B) 12 Pa
- (C) 1.2 × 104 Pa
- (D) 3.0 × 105 Pa
Show Answer
Answer: A – p = F / A = 600 / 0.020 = 30 000 Pa = 3.0 × 104 Pa. Option B is F × A – a common slip.
Q12. The pressure at a depth h in a liquid is given by p = ρgh, obtained by dividing the weight of the liquid column above the point by its cross-sectional area. Which quantity does the pressure at a point in a static liquid NOT depend on?
- (A) The density of the liquid
- (B) The cross-sectional area of the container
- (C) The depth below the surface
- (D) The gravitational field strength
Show Answer
Answer: B – In the derivation, the mass of the column is ρAh, so its weight is ρAhg; dividing by A cancels the area, leaving p = ρgh. Pressure at a depth depends only on ρ, g and h – not on the shape or area of the container.
Q13. A diver is 20 m below the surface of the sea. Sea water has density 1030 kg m-3. What is the pressure due to the water alone at this depth?
- (A) 2.0 × 104 Pa
- (B) 2.0 × 106 Pa
- (C) 2.0 × 105 Pa
- (D) 1.0 × 105 Pa
Show Answer
Answer: C – p = ρgh = 1030 × 9.8 × 20 = 2.02 × 105 Pa ≈ 2.0 × 105 Pa. This is the gauge pressure – the excess over atmospheric pressure.
Q14. Atmospheric pressure is 1.0 × 105 Pa. Using your answer to the diver situation (water pressure 2.0 × 105 Pa at 20 m depth), what is the total (absolute) pressure on the diver?
- (A) 1.0 × 105 Pa
- (B) 2.0 × 105 Pa
- (C) 4.0 × 105 Pa
- (D) 3.0 × 105 Pa
Show Answer
Answer: D – The atmosphere presses on the sea surface and this pressure is transmitted through the water, so absolute pressure = atmospheric + hydrostatic = 1.0 × 105 + 2.0 × 105 = 3.0 × 105 Pa.
Q15. In a continuous liquid at rest, two points X and Y lie at the same horizontal level. How do the pressures at X and Y compare?
- (A) They are equal, because pressure in a static liquid depends only on depth
- (B) The pressure is greater at the point nearer the container wall
- (C) The pressure is greater at the point under the larger volume of liquid
- (D) They are equal only if the container is cylindrical
Show Answer
Answer: A – Since p = ρgh and both points are at the same depth h in the same liquid, the pressures are equal. This is why connected containers of any shape fill to the same level, and it is the basis of the liquid manometer.
Q16. A water manometer measures the pressure of a gas supply. The difference in the water levels in the two arms is 0.20 m. By how much does the gas pressure exceed atmospheric pressure? (Density of water = 1000 kg m-3.)
- (A) 2.0 × 102 Pa
- (B) 2.0 × 103 Pa
- (C) 2.0 × 104 Pa
- (D) 196 Pa
Show Answer
Answer: B – The excess pressure supports the extra column of water of height 0.20 m: Δp = ρgh = 1000 × 9.8 × 0.20 = 1960 Pa ≈ 2.0 × 103 Pa. Option D is the unrounded value of ρh without g.
Q17. A manometer containing mercury (density 13 600 kg m-3) shows a height difference of 0.12 m. What pressure difference does this represent?
- (A) 1.6 × 103 Pa
- (B) 1.6 × 105 Pa
- (C) 1.6 × 104 Pa
- (D) 1.2 × 104 Pa
Show Answer
Answer: C – Δp = ρgh = 13 600 × 9.8 × 0.12 = 1.6 × 104 Pa. Mercury’s high density keeps the column short: the same pressure difference would need a water column of about 1.6 m.
Q18. In a hydraulic press, a force of 40 N is applied to a small piston of area 5.0 cm2. The large piston has area 200 cm2. What force is exerted by the large piston?
- (A) 40 N
- (B) 400 N
- (C) 1000 N
- (D) 1600 N
Show Answer
Answer: D – The pressure is transmitted equally: p = F / A = 40 / (5.0 × 10-4) = 8.0 × 104 Pa. Force on the large piston = pA = 8.0 × 104 × (200 × 10-4) = 1600 N. Equivalently, the force is multiplied by the area ratio 200 / 5.0 = 40.
Q19. Which statement about pressure at a point inside a static liquid is correct?
- (A) It acts only in the downward direction
- (B) It acts equally in all directions
- (C) It is zero in the horizontal direction
- (D) It increases in the direction away from the container walls
Show Answer
Answer: B – Pressure in a fluid at rest is isotropic: at a given point it has the same value in every direction. Its magnitude increases with depth, but at the point itself it pushes equally on a surface facing any way – which is why a submerged object experiences forces on all its faces.
Q20. A reservoir dam holds back water of depth 30 m (density 1000 kg m-3). What is the pressure due to the water at the base of the dam?
- (A) 2.9 × 105 Pa
- (B) 2.9 × 104 Pa
- (C) 3.0 × 103 Pa
- (D) 2.9 × 106 Pa
Show Answer
Answer: A – p = ρgh = 1000 × 9.8 × 30 = 2.94 × 105 Pa ≈ 2.9 × 105 Pa. Because pressure grows with depth, dams are built thicker at the base.
Upthrust & Floating
Q21. According to Archimedes’ principle, the upthrust on a body immersed in a fluid is equal to
- (A) the weight of the fluid displaced by the body
- (B) the weight of the body
- (C) the mass of the fluid displaced by the body
- (D) the volume of the body multiplied by g
Show Answer
Answer: A – Upthrust equals the weight of the fluid displaced: U = mfluidg = ρgV, where V is the volume of fluid displaced. Option C gives a mass, not a force.
Q22. A block of volume 2.0 × 10-3 m3 is held completely immersed in water of density 1000 kg m-3. What is the upthrust on the block?
- (A) 1.96 N
- (B) 196 N
- (C) 19.6 N
- (D) 0.0196 N
Show Answer
Answer: C – U = ρgV = 1000 × 9.8 × 2.0 × 10-3 = 19.6 N. The block’s own material is irrelevant – only the displaced water matters.
Q23. A large block floats with exactly half of its volume below the water surface. The upthrust on the block equals
- (A) the weight of a volume of water equal to the block’s total volume
- (B) the weight of a volume of water equal to the submerged half of the block
- (C) half the weight of the block
- (D) the weight of the block multiplied by two
Show Answer
Answer: B – Only the submerged part displaces water, so V in U = ρgV is the submerged volume. For a floating body this upthrust exactly balances the block’s full weight, even though only half of it is underwater.
Q24. Ice has density 920 kg m-3. What fraction of a floating iceberg’s volume lies below the surface of sea water of density 1000 kg m-3?
- (A) 0.08
- (B) 0.50
- (C) 1.00
- (D) 0.92
Show Answer
Answer: D – For a floating body, upthrust = weight: ρwater g Vsub = ρice g Vtotal, so the fraction submerged = ρice / ρwater = 920 / 1000 = 0.92. Hence 92% of an iceberg is hidden underwater.
Q25. A wooden block of density 400 kg m-3 floats in water of density 1000 kg m-3. What percentage of the block’s volume is submerged?
- (A) 40%
- (B) 60%
- (C) 25%
- (D) 4%
Show Answer
Answer: A – Fraction submerged = ρblock / ρwater = 400 / 1000 = 0.40, so 40% of the volume is below the surface and 60% is above it.
Q26. Steel has density 7800 kg m-3, yet a steel ship floats. What is the best explanation?
- (A) The steel used for ships has a density lower than water
- (B) The upthrust on a floating ship is greater than its weight
- (C) The average density of the ship, including the air inside its hollow hull, is less than the density of water
- (D) The ship displaces a weight of water smaller than its own weight
Show Answer
Answer: C – The hull encloses a large volume of air, so the ship’s total mass divided by its total volume (its average density) is less than 1000 kg m-3. It sinks only until the weight of water displaced equals its total weight. Option B is wrong: for any body floating in equilibrium, upthrust exactly equals weight.
Q27. A stone of weight 50 N hangs from a spring balance. When the stone is completely immersed in water, the upthrust on it is 20 N. What does the balance read?
- (A) 50 N
- (B) 70 N
- (C) 20 N
- (D) 30 N
Show Answer
Answer: D – The balance reads the apparent weight = actual weight − upthrust = 50 − 20 = 30 N. The “loss in weight” of an immersed object is exactly equal to the upthrust.
Q28. A hydrometer is placed first in water and then in a liquid denser than water. In the denser liquid, the hydrometer
- (A) sinks lower, with more of the stem submerged
- (B) floats higher, with less of the stem submerged
- (C) sinks to the bottom
- (D) floats at exactly the same level
Show Answer
Answer: B – The hydrometer’s weight is fixed, so it needs to displace the same weight of liquid in each case. In a denser liquid a smaller submerged volume provides that weight of liquid, so the hydrometer rises. This is why hydrometer scales read increasing density downwards along the stem.
Q29. A metal block of mass 0.50 kg and volume 5.0 × 10-4 m3 is suspended by a thread and completely immersed in oil of density 800 kg m-3. What is the tension in the thread?
- (A) 1.0 N
- (B) 3.9 N
- (C) 4.9 N
- (D) 8.8 N
Show Answer
Answer: A – Weight = mg = 0.50 × 9.8 = 4.9 N. Upthrust = ρgV = 800 × 9.8 × 5.0 × 10-4 = 3.92 N. Tension = weight − upthrust = 4.9 − 3.92 = 0.98 N ≈ 1.0 N. Option B is the upthrust alone; option D adds instead of subtracting.
Q30. A block of mass 2.0 kg floats in water of density 1000 kg m-3. What volume of the block is below the water surface?
- (A) 2.0 m3
- (B) 2.0 × 10-2 m3
- (C) 2.0 × 10-3 m3
- (D) 2.0 × 10-1 m3
Show Answer
Answer: C – Floating means upthrust = weight, so ρgVsub = mg and Vsub = m / ρ = 2.0 / 1000 = 2.0 × 10-3 m3. Note that g cancels – the submerged volume depends only on the mass and the liquid’s density.
All questions are original practice questions written for QuaidianPoint – not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.