The short answer
Given a vessel, a floating or submerged object, or a pipe that changes width, decide which principle governs it and predict the pressure, force or speed before reaching for a formula.
Written and checked by GAMSAT tutors — not AI-generated.
Try the reasoning style
We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.
The author's argument relies most directly on which unstated assumption?
Pick an option to see how the tutor reasons to the answer — not just whether you were right.
Not quite — the answer is B.
Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.
Fluids hides physics in a costume — a stenosed artery, a dam wall. Underneath sit three questions: how hard is the fluid pushing, what holds this object up, how fast is it moving? Name which one first.
Pushing. In P = P₀ + ρgh, ρgh is the weight of fluid above you per unit area: shape and volume never appear, so equal depths in one connected fluid share a pressure.
| Vessel | Shape | Water depth (m) | Area of the base (m²) | Water held (L) | Gauge pressure at the base (kPa) |
|---|---|---|---|---|---|
| A | straight-sided tube | 1.5 | 0.002 | 3 | 15 |
| B | straight-sided drum | 1.5 | 0.200 | 300 | 15 |
| C | narrow at the base, flaring to a wide bowl | 1.5 | 0.002 | 100 | 15 |
Floating versus fully submerged
Floating: displaces its own WEIGHT
- Buoyancy equals the object's weight — equilibrium, not a maximum
- Fraction submerged = ρ_object/ρ_fluid
- Denser fluid, rides higher
Fully submerged: displaces its own VOLUME
- F_b = ρ_fluid V g — volume alone
- Depth changes nothing — both faces gain equally
- Equal-sized balsa and lead, both held under, feel identical buoyancy
Worked example — boat and nut
A toy boat carries a steel nut, mass 40 g, volume 5 cm³. Dropped overboard, it sinks to the bottom. Does the tank's water level rise, fall, or stay the same?
Flow: continuity first, Bernoulli second
Order matters
A₁v₁ = A₂v₂, and area goes as r², so at fixed flow v ∝ 1/r² — halve the radius, quadruple the speed. Bernoulli comes second — P + ½ρv² + ρgh constant along a steady, non-viscous streamline — so where speed rose, pressure fell: fast flow is low pressure.
The draining tank
Surface and hole are both open to air, so the P terms cancel: ρgh = ½ρv², giving v = √(2gh). Quadruple the depth to double the jet.
- speed
- pressure
Which power? Ask what is held constant
Flow rate fixed through a narrowing is continuity: speed rises as 1/r², and pressure there falls. Driving pressure fixed across a long viscous tube is resistance: flow scales as r⁴, so a 20% narrowing passes 0.8⁴ ≈ 41%. Viscosity or turbulence: Bernoulli no longer applies.
Check yourself
A sealed hollow float of mass 150 g and external volume 500 cm³ is released in a tank of fresh water (density 1.00 g cm⁻³) and settles at the surface. A short vertical cable is then attached from the float to the floor of the tank, holding it at rest completely below the surface. Take g = 10 m s⁻². What is the tension in the cable?
Key takeaways
- Depth alone sets pressure in a still fluid — force also needs area.
- Floating displaces its own weight; submerged, its own volume — F_b = ρ_fluid V g, whatever the depth or material.
- Continuity fixes speed before Bernoulli fixes pressure: faster means lower pressure.
- Three powers, three questions: 1/r² at fixed flow, r⁴ at fixed driving pressure, √h for a jet.
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