Why can a hydraulic press lift heavy loads?

Pressure is the same throughout the enclosed liquid, so the larger piston gets a force scaled by its area.

How do floating and sinking differ?

A body floats when the upthrust can equal its weight; a floating body has upthrust equal to weight, a sinking one less.

S4 Physics pressure and fluids paper 2: hydraulics and upthrust

2026-11-10

Tutor Prep Guide · S4 Physics Pressure and Fluids

Ten items apply the pressure formulas to presses, floating bodies and apparent weight, all checked in python.

Direct answers　Items 01 to 03 cover presses and solids, 04 to 06 upthrust and floating, 07 to 09 mix depth, volume and ratio. Allow 25 minutes.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Physics Pressure and Fluids
Level	S4

02　Questions (10 items)

FoundationA hydraulic press has a small piston of 0.002 m squared under 100 N and a large piston of 0.5 m squared. Find the force it can raise.
FoundationOil of density 800 kg per cubic metre is 2.5 m deep. Find the pressure at the bottom (g = 10 N/kg).
FoundationA cube of side 0.2 m with mass 8 kg rests flat on a table. Find the pressure (g = 10 N/kg).
IntermediateAn object of volume 0.002 m cubed is fully immersed in water. Find the upthrust (density 1000, g = 10).
IntermediateA 12 N block floats on water. Find the upthrust and the volume of water displaced.
IntermediateTwo points in a pipe differ in height by 0.4 m. Find the pressure difference (density 1000, g = 10).
ChallengeA diver is at 25 m depth. Find the total pressure and express it in atmospheres (1 atm = 100000 Pa).
ChallengeA small piston of area 0.002 m squared moves down 0.2 m. The large piston has area 0.5 m squared. How far does it rise?
ChallengeIce of density 900 kg per cubic metre and volume 0.01 m cubed floats on water. Find the submerged volume.
Error redoA pupil says the deeper an object is, the larger the upthrust. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	25000 N	The pressure is the same: 100 / 0.002 = 50000 Pa, then times 0.5 m squared gives 25000 N.
2	20000 Pa	P = rho x g x h = 800 x 10 x 2.5 = 20000 Pa; oil is less dense than water.
3	2000 Pa	Weight is 80 N and the contact area is 0.2 x 0.2 = 0.04 m squared, so P = 80 / 0.04 = 2000 Pa.
4	20 N	Upthrust = rho x g x displaced volume = 1000 x 10 x 0.002 = 20 N.
5	12 N; 0.0012 m cubed	A floating body has upthrust equal to its weight; use F = rho g V to find the displaced volume.
6	4000 Pa	Only the height difference counts: 1000 x 10 x 0.4 = 4000 Pa.
7	350000 Pa, that is 3.5 atmospheres	Water pressure is 1000 x 10 x 25 = 250000 Pa, and adding 100000 Pa gives 350000 Pa.
8	0.0008 m	The liquid volume is unchanged: 0.002 x 0.2 = 0.0004 m cubed, then divide by 0.5 m squared.
9	0.009 m cubed, that is nine tenths	The ice weighs 90 N, so the upthrust is 90 N and the displaced volume is 90 / 10000 = 0.009 m cubed.
10	Once fully submerged, upthrust = rho x g x displaced volume, which depends on the liquid and the volume, not the depth.	Depth changes pressure but not the displaced volume, so the upthrust stays the same.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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