These eight questions move from one-step pressure to reading a hydraulic table. Try each on paper before you open the answer.
They cover the cluster on force and pressure. The timed practice session builder can turn the set into a timed run.
Questions
Question 1 (2 marks). A force of 60 N acts on an area of 0.30 m². Calculate the pressure.
Answer
Pressure = force ÷ area = 60 ÷ 0.30 = 200 Pa. Write the formula, the substitution and the unit for full marks.
Question 2 (3 marks). A box weighs 120 N. It rests on a face of area 0.06 m², then is turned to rest on a face of area 0.02 m². Calculate the pressure each time and state which is larger.
Answer
First face: 120 ÷ 0.06 = 2 000 Pa. Second face: 120 ÷ 0.02 = 6 000 Pa.
The pressure is larger on the smaller face. The weight is unchanged, so a smaller area gives a larger pressure.
Question 3 (2 marks). A dam wall is thicker at the bottom than at the top. Explain why.
Answer
Liquid pressure increases with depth. The water at the bottom pushes on the wall harder, so the wall must be thicker there to withstand the larger pressure.
Question 4 (3 marks). A student measures the pressure in water at three depths.
| Depth (m) | 0.2 | 0.4 | 0.6 |
|---|---|---|---|
| Pressure (Pa) | 2 000 | 4 000 | 6 000 |
State the relationship, and predict the pressure at 0.8 m.
Answer
The pressure is directly proportional to depth: each extra 0.2 m adds 2 000 Pa.
At 0.8 m, the pressure is 8 000 Pa. The prediction assumes the same liquid and the same conditions.
Question 5 (2 marks). A wooden block of mass 0.5 kg floats still on water. Take the weight of 1 kg as 10 N. State the buoyant force and give a reason.
Answer
The weight is 0.5 × 10 = 5 N. The buoyant force is 5 N, upwards.
The block is not moving, so the forces are balanced. The upward buoyant force must equal the downward weight.
Question 6 (3 marks). A hydraulic press has a small piston of area 3 cm² and a large piston of area 60 cm². Calculate the ideal output force for an input force of 25 N.
Answer
The area ratio is 60 ÷ 3 = 20. Pressure under the small piston = 25 ÷ 3 N/cm², so the output is (25 ÷ 3) × 60 = 500 N.
The shortcut gives the same result: 25 × 20 = 500 N.
Question 7 (4 marks). The same press is tested and the measured output is 440 N for a 25 N input. Calculate the percentage of the ideal output that was delivered, and give one reason for the difference.
Answer
Percentage = 440 ÷ 500 × 100 = 88%.
Reason: friction between the pistons and the cylinder wastes some input force. An air bubble that compresses, or a small leak, gives the same kind of loss.
Question 8 (4 marks). A student claims: “A hydraulic jack gives out more energy than I put in, because the force is larger.” Evaluate the claim.
Answer
The claim is wrong. The output force is larger, but the large piston moves a shorter distance than the small piston.
Work is force × distance, so the output work cannot exceed the input work. In a real jack, friction makes the output work slightly smaller still.
If you got these wrong
- Questions 1 to 3: revisit calculating pressure from force and area.
- Questions 4 and 5: read comparing pressure in fluids and explaining buoyancy conceptually.
- Questions 6 to 8: work through interpreting hydraulic system data.
A mistake log helps you see which step keeps repeating. If you want a teacher to read your wording, see online one-to-one Science tuition.