Pressure in liquids — answer key

1. PRQ13 · Practice

Model answer. Only the depth below the surface and the density of the liquid set the pressure. Row A: the same depth, but glycerol is denser than water, so the pressure is greater at P, not the same at both. Row B: the same depth, and water is denser than olive oil, so P is greater; the width of a tank makes no difference. Row C: Q is deeper and in a denser liquid, so the pressure is greater at Q. Row D: depths are measured down from the surface, so P is 0.95 m deep and Q only 0.25 m deep; P is greater. Row C is correct.

A full answer includes

• I chose row C.
• I did not measure a depth up from the bottom of a tank.

Also accept

• Row C

Watch for

• “Row A.” — Glycerol is denser than water. At the same depth, the denser liquid gives the greater pressure, so P is greater.
• “Row B.” — The width of a tank does not change the pressure at a depth. Water is denser than olive oil, so P is greater.
• “Row D.” — Depth is measured down from the surface, not up from the bottom. P is 0.95 m deep and Q only 0.25 m, so P is greater.

2. PRQ14 · Practice

Model answer. Both sensors are at the same depth, and both jars are open to the same air. At the same depth, the pressure is greater in a denser liquid, because each cubic metre of it has more mass, and so more weight, pressing down. So liquid X is denser than liquid Y.

A full answer includes

• I chose: liquid X is denser than liquid Y.
• I did not choose that liquid X is less dense than liquid Y.

Also accept

• liquid X is denser than liquid Y

Watch for

• “Liquid X is less dense than liquid Y.” — A less dense liquid gives a smaller pressure at the same depth, not a larger one.
• “Liquid X flows more easily than liquid Y.” — How easily a liquid flows does not set its pressure. At the same depth, the density decides it.
• “The air above jar X presses harder.” — Both jars stand open in the same room, so the air presses equally on both surfaces. The difference comes from the liquids.

3. PRQ15 · Practice

Options as numbered on the pupil sheet: 1 — The reading stays the same.; 2 — The reading decreases.; 3 — The reading increases..

Model answer. The pressure at the sensor depends only on its depth below the surface and on the density of the liquid. Pouring in more water raises the surface, so the sensor is deeper below it: the reading increases. Raising the sensor makes it less deep: the reading decreases. Sliding it towards a wall at the same depth changes neither the depth nor the liquid, so the reading stays the same. At any depth, the water pushes equally in all directions, so turning the sensor to face upwards does not change the reading either.

A full answer includes

• I matched pouring in more water to: the reading increases.
• I matched raising the sensor to: the reading decreases.
• I matched sliding it towards a wall to: the reading stays the same.
• I matched turning it to face upwards to: the reading stays the same.
• I did not match pouring in more water to: the reading stays the same.

Also accept

• more water – increases; raised – decreases; towards a wall – stays the same; facing upwards – stays the same

Watch for

• “The sensor has not moved, so pouring in water leaves the reading the same.” — Depth is measured down from the surface. When the surface rises, the sensor is deeper below it, so the reading increases.
• “Water only presses down, so facing upwards the sensor reads less.” — At any point, a liquid pushes equally in all directions, up as well as down.

4. PRQ16 · Practice

Options as numbered on the pupil sheet: 1 — 0.65 m below the surface of water; 2 — 0.12 m below the surface of ethanol; 3 — 0.65 m below the surface of sugar solution; 4 — 0.12 m below the surface of water.

Model answer. At 0.12 m, ethanol is less dense than water, so its pressure is smaller. In water, 0.65 m is deeper than 0.12 m, so its pressure is greater. At 0.65 m, sugar solution is denser than water, so its pressure is the greatest. Order: 0.12 m in ethanol, 0.12 m in water, 0.65 m in water, 0.65 m in sugar solution.

A full answer includes

• I put 0.12 m in ethanol first.
• I put 0.12 m in water second.
• I put 0.65 m in water third.
• I put 0.65 m in sugar solution last.
• I did not put water at 0.12 m before ethanol at 0.12 m.

Also accept

• 0.12 m in ethanol, 0.12 m in water, 0.65 m in water, 0.65 m in sugar solution

Watch for

• “Only the depth matters, so the two 0.12 m points are equal.” — The density matters too: at the same depth, the denser liquid gives the greater pressure.

5. PRQ17 · Practice

Model answer. (a) At any point in a liquid, the pressure pushes equally in all directions: down, sideways and up. So the water under the boat pushes up on the floor window. (b) They are the same. Both points are 0.70 m below the surface of the same water, and the direction the window faces does not matter. (c) It stays the same. The pressure depends only on the depth and the density of the water, not on how much water there is around the boat.

A full answer includes

• (a) I said that at any point in a liquid, the pressure acts equally in all directions.
• (a) I said that this includes upwards, onto the floor window.
• (b) I wrote that the two pressures are the same.
• (b) I said both points are 0.70 m below the surface of the same water.
• (c) I wrote that the pressure stays the same.
• (c) I said the amount of water around the boat does not change the pressure.
• I did not write that water only pushes downwards.

Also accept

• (a) pressure acts in all directions, including up (b) the same (c) it stays the same

Watch for

• “Water only pushes down, so it cannot push up on the floor window.” — At any point, a liquid pushes equally in all directions: up, down and sideways.
• “The open sea holds far more water, so the pressure on the window is greater.” — At the same depth in the same water, the pressure is the same, however much water there is around the boat.

6. PRQ18 · Practice

Model answer. From the surface down to the bottom, ∆h = 4.5 m. ∆p = ρg∆h = 1000 kg/m³ × 9.8 N/kg × 4.5 m = 44 100 Pa, or 44.1 kPa.

Checked automatically

• Pressure of the water: 44100 Pa (within ±500 Pa)
  • Common wrong answer 4500 : Multiply by g, 9.8 N/kg, as well: ∆p = ρg∆h.
  • Common wrong answer 45000 : The question gives g = 9.8 N/kg, not 10 N/kg.

A full answer includes

• Method: I multiplied 1000 kg/m³ × 9.8 N/kg × 4.5 m.
• Answer: I wrote 44 100 Pa (or 44 000 Pa).
• I did not leave out g.

Also accept

• 44 100 Pa

Watch for

• “∆p = 1000 × 4.5 = 4500 Pa.” — Multiply by g, 9.8 N/kg, as well.
• “∆p = 1000 × 10 × 4.5 = 45 000 Pa.” — The question gives g = 9.8 N/kg. Use that value.

7. PRQ19 · Practice

Model answer. The change in depth is ∆h = 43 m − 16 m = 27 m. ∆p = ρg∆h = 1000 kg/m³ × 9.8 N/kg × 27 m = 264 600 Pa ≈ 265 000 Pa to 3 significant figures.

Checked automatically

• Increase in pressure: 2.65×10^5 Pa (3 significant figures)
  • Common wrong answer 421400 : ∆h is the change in depth, 43 m − 16 m = 27 m, not 43 m.
  • Common wrong answer 27000 : Multiply by g, 9.8 N/kg, as well: ∆p = ρg∆h.
  • Common wrong answer 270000 : The question gives g = 9.8 N/kg, not 10 N/kg.

A full answer includes

• Method: I worked out ∆h = 43 m − 16 m = 27 m.
• Method: I multiplied 1000 kg/m³ × 9.8 N/kg × 27 m.
• Answer: I wrote 265 000 Pa, to 3 significant figures.
• I did not use 43 m as ∆h.

Also accept

• 265 000 Pa to 3 significant figures (264 600 Pa)

Watch for

• “∆h = 43 m.” — ∆h is the change in depth: the submarine starts at 16 m, so ∆h = 27 m.

8. PRQ20 · Practice

Model answer. Rearrange ∆p = ρg∆h: ∆h = ∆p ÷ (ρg) = 20 580 Pa ÷ (840 kg/m³ × 9.8 N/kg) = 20 580 ÷ 8232 = 2.5 m.

Checked automatically

• Depth: 2.5 m (within ±0.02 m)
  • Common wrong answer 24.5 : Divide by g, 9.8 N/kg, as well: ∆h = ∆p ÷ (ρg).
  • Common wrong answer 2.1 : That uses the density of water. Use the density of the diesel, 840 kg/m³.
  • Common wrong answer 2.45 : The question gives g = 9.8 N/kg, not 10 N/kg.

A full answer includes

• Method: I rearranged ∆p = ρg∆h to ∆h = ∆p ÷ (ρg).
• Method: I divided 20 580 Pa by 840 kg/m³ × 9.8 N/kg.
• Answer: I wrote 2.5 m.
• I did not use the density of water.

Also accept

• 2.5 m

Watch for

• “∆h = 20 580 ÷ 840 = 24.5 m.” — Divide by g as well: ∆h = ∆p ÷ (ρg).

9. PRQ21 · Practice

Model answer. Any point on the line will do, for example 16.0 kPa at 2.0 m. ∆p = 16.0 kPa = 16 000 Pa for ∆h = 2.0 m. ρ = ∆p ÷ (g∆h) = 16 000 Pa ÷ (9.8 N/kg × 2.0 m) = 816.3… kg/m³ ≈ 820 kg/m³ to 2 significant figures. Liquid X is less dense than water.

Checked automatically

• Density of X: 8.2×10^2 kg/m³ (2 significant figures)
  • Common wrong answer 0.82 : Change kPa into Pa first: 16.0 kPa = 16 000 Pa.
  • Common wrong answer 8000 : Divide by g, 9.8 N/kg, as well.
  • Common wrong answer 800 : The question gives g = 9.8 N/kg, not 10 N/kg.

A full answer includes

• Method: I read one point on the line, such as 16.0 kPa at 2.0 m.
• Method: I changed 16.0 kPa into 16 000 Pa.
• Method: I divided 16 000 Pa by (9.8 N/kg × 2.0 m).
• Answer: I wrote 820 kg/m³, to 2 significant figures.
• I did not leave the pressure in kPa.

Also accept

• 820 kg/m³ to 2 significant figures (816 kg/m³)

Watch for

• “ρ = 16.0 ÷ (9.8 × 2.0) = 0.82 kg/m³.” — Change kPa into Pa first: 16.0 kPa = 16 000 Pa.
• “ρ = 16 000 ÷ (10 × 2.0) = 800 kg/m³.” — The question gives g = 9.8 N/kg. Use that value.

10. PRQ22 · Practice

Model answer. ∆h is the change in depth: 9.0 m − 4.0 m = 5.0 m. The depth of the lake below the camera does not matter.

A full answer includes

• I chose 5.0 m.
• I did not measure from the bottom of the lake.

Also accept

• 5.0 m

Watch for

• “∆h = 9.0 m.” — 9.0 m is where the camera started. ∆h is how much its depth changed.
• “∆h = 11 m.” — That is 20 m − 9.0 m, a height above the bottom of the lake. Depths are measured down from the surface.
• “∆h = 13 m.” — That adds the two depths. ∆h is the difference between them.

11. PRQ23 · Practice

Model answer. ∆p = ρg∆h, and g is the same for every row, so compare ρ × ∆h. A: 740 × 2.4 = 1776. B: 1420 × 1.2 = 1704. C: 1000 × 1.9 = 1900. D: 1035 × 1.7 = 1759.5. Row C is the greatest: 1000 kg/m³ × 9.8 N/kg × 1.9 m = 18 620 Pa. The water is neither the densest liquid nor at the deepest point, but its product of density and depth is the largest.

A full answer includes

• Method: I worked out density × depth for every row.
• Answer: I chose row C.
• I did not choose a row only because its liquid is the densest or its point the deepest.

Also accept

• Row C

Watch for

• “Row A.” — This point is the deepest, but petrol is the least dense: 740 × 2.4 = 1776, less than water’s 1000 × 1.9 = 1900.
• “Row B.” — Honey is the densest, but this point is the shallowest: 1420 × 1.2 = 1704, less than water’s 1900.
• “Row D.” — Milk is denser than water, but this point is less deep: 1035 × 1.7 = 1759.5, less than water’s 1900.

12. PRT2 · Apply

Model answer. (a) The pressure in a liquid increases with depth. The ground-floor tap is farther below the water surface in the tank than the top-floor tap, so the pressure there is greater. (b) ∆p = ρg∆h = 1000 kg/m³ × 9.8 N/kg × 30 m = 294 000 Pa. (c) The taps differ in depth by ∆h = 30 m − 24 m = 6.0 m, so ∆p = 1000 kg/m³ × 9.8 N/kg × 6.0 m = 58 800 Pa.

A full answer includes

• (a) I said the ground-floor tap is deeper below the water surface in the tank.
• (a) I said a greater depth gives a greater pressure.
• (a) I did not write that the higher tap has more pressure.
• (b) Method: I multiplied 1000 kg/m³ × 9.8 N/kg × 30 m.
• (b) Answer: I wrote 294 000 Pa.
• (c) Method: I worked out ∆h = 30 m − 24 m = 6.0 m.
• (c) Method: I multiplied 1000 kg/m³ × 9.8 N/kg × 6.0 m.
• (c) Answer: I wrote 58 800 Pa.
• (c) I did not use 24 m as ∆h.

Also accept

• (a) deeper below the surface, so greater pressure (b) 294 000 Pa (c) 58 800 Pa

Watch for

• “The top floor is higher, so it has more pressure.” — Pressure grows with depth below the water surface; the top floor is less deep.
• “In (c), use 24 m.” — The difference in pressure needs the difference in depth, 6.0 m.