The problems are on the density and pressure worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Density is mass per unit volume, ρ = m ÷ V. These worked answers correct the 20 problems of the Physics 9-10 worksheet on density and pressure, from formulas and units to submarine data: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the equation, the substitution and the result with its unit in bold. Check that you converted units first and used g = 9.8 N/kg. If your final value differs, compare each line of working with the model to find where the difference starts.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P3, IGCSE 1.8
Problem 1 Formulas and units
1. ρ = m ÷ V.2. p = F ÷ A.
3. 2.7 × 1000 = 2700 kg/m³.
4. The pascal: 1 Pa = 1 N/m².
Problem 2 Wooden plank
1. ρ = 3.6 ÷ 0.0050 = 720 kg/m³.2. 720 < 1000, so it floats.
Problem 3 Petrol tanker
1. m = ρ × V.2. m = 740 × 25 = 18 500 kg.
Problem 4 Volume of mercury
1. V = m ÷ ρ.2. V = 0.50 ÷ 13 600 = 3.7 × 10⁻⁵ m³.
3. 3.68 × 10⁻⁵ × 1 000 000 = 37 cm³.
Problem 5 Float or sink?
1. Floats: 520 < 1000.2. Sinks: 2100 > 1000.
3. Floats: 920 < 1000.
4. Sinks: 1260 > 1000.
Problem 6 Pressure of a press
1. p = F ÷ A = 300 ÷ 0.0020 = 150 000 Pa.2. 150 kPa.
Problem 7 Ore sample
1. 50.0 − 40.0 = 10.0 cm³.2. ρ = 84.5 ÷ 10.0 = 8.45 g/cm³.
3. 8.45 × 1000 = 8450 kg/m³.
Problem 8 Steel bar
1. V = 0.50 × 0.040 × 0.025 = 5.0 × 10⁻⁴ m³.2. m = ρ V = 7800 × 5.0 × 10⁻⁴ = 3.9 kg.
Problem 9 Tractor tyres
1. W = 4500 × 9.8 = 44 100 N.2. p = 44 100 ÷ 0.60 = 73 500 Pa (73.5 kPa).
3. p = 44 100 ÷ 0.20 = 220 500 Pa (three times larger).
Problem 10 Drawing pin
1. p = 20 ÷ 1.0 × 10⁻⁴ = 2.0 × 10⁵ Pa.2. p = 20 ÷ 1.0 × 10⁻⁷ = 2.0 × 10⁸ Pa.
3. The same force acts on a tip area 1000 times smaller, so the pressure is 1000 times greater and is enough to push into the board.
Problem 11 Rearranging p = F ÷ A
1. A = F ÷ p = 800 ÷ 40 000 = 0.020 m².2. F = p A = 25 000 × 0.60 = 15 000 N.
Problem 12 Water tower
1. p = ρ g h = 1000 × 9.8 × 25 = 245 000 Pa.2. 245 kPa.
Problem 13 Pressure-depth graph
1. 29.4 kPa.2. Gradient = 39.2 ÷ 4.0 = 9.8 kPa/m.
3. Gradient = ρ g, so ρ = 9800 ÷ 9.8 = 1000 kg/m³.
Problem 14 Sluice gate
1. p = 1000 × 9.8 × 30 = 294 000 Pa.2. A = 0.50 × 0.40 = 0.20 m².
3. F = p A = 294 000 × 0.20 = 58 800 N.
Problem 15 Sea and river
1. p = 1000 × 9.8 × 5.0 = 49 000 Pa.2. p = 1030 × 9.8 × 5.0 = 50 470 Pa.
3. 50 470 − 49 000 = 1470 Pa.
Problem 16 Model steel boat
1. ρ = 0.80 ÷ 1.2 × 10⁻³ = 670 kg/m³ (667).2. 667 < 1000, so it floats.
3. V = 0.80 ÷ 7800 = 1.0 × 10⁻⁴ m³ (103 cm³).
4. A solid block has the density of steel, 7800 kg/m³, which is greater than that of water. The boat contains a lot of air, which lowers its average density.
Problem 17 Brass key
1. V = 60 ÷ 8.9 = 6.74 cm³.2. V = 40 ÷ 7.1 = 5.63 cm³.
3. ρ = 100 ÷ (6.74 + 5.63) = 100 ÷ 12.38 = 8.1 g/cm³.
4. It is an average of the two densities, weighted by the volumes. It cannot be greater than the densest component or less than the lightest.
Problem 18 Diver at 18 m
1. p = 1030 × 9.8 × 18 = 181 692 Pa, so 1.8 × 10⁵ Pa.2. Total = 181 692 + 101 000 = 282 692, so 2.8 × 10⁵ Pa.
3. F = 181 692 × 0.015 = 2725 N, so 2.7 × 10³ N (about 2700 N) from the water; air inside the mask pushes back from the other side.
Problem 19 Block on its faces
1. V = 0.50 × 0.20 × 0.10 = 0.010 m³; ρ = 12 ÷ 0.010 = 1200 kg/m³.2. W = 12 × 9.8 = 117.6 N.
3. Smallest face: 0.20 × 0.10 = 0.020 m², p = 117.6 ÷ 0.020 = 5880 Pa. Largest face: 0.50 × 0.20 = 0.10 m², p = 117.6 ÷ 0.10 = 1176 Pa.
4. The greatest pressure is on the 0.20 m × 0.10 m face and the smallest on the 0.50 m × 0.20 m face.
Problem 20 Submarine data
1. ρ = p ÷ (g h): 101 000 ÷ 98 = 1031; 203 000 ÷ 196 = 1036; 303 000 ÷ 294 = 1031; 405 000 ÷ 392 = 1033 (kg/m³).2. Mean = (1031 + 1036 + 1031 + 1033) ÷ 4 = 1033 kg/m³, close to the accepted 1030 kg/m³.
3. p = 1033 × 9.8 × 55 = 556 800 Pa, so about 5.6 × 10⁵ Pa (557 kPa).
