Physics of liquids

What the exam asks

Expect a diagram: a vessel of an odd shape, a U-tube, a block floating with some fraction showing, or an object on a balance weighed in air and then in water. The formulae are printed. The items are: compare the pressure at two marked points, read a height difference off a manometer, get a density from a floating fraction, or get an upthrust from the gap between two balance readings. The trap that costs the most is reasoning from how much liquid there is. A narrow tube and a wide tank filled to the same depth have the same pressure at the bottom, and a candidate who thinks the tank has more water pushing down loses every hydrostatic paradox item there is. Read the depth, not the volume. The second trap is putting the wrong volume into rho V g: check whether the object floats or is fully under first, because the arithmetic is trivial and that decision is not.

This topic is hydrostatics: how pressure behaves in a liquid at rest, and what happens to an object put into one. There are two ideas in it, and the exam gets a great deal of mileage out of both, usually with a diagram of an awkwardly shaped vessel or a floating block and a formula printed in the stem.

The first idea is that pressure depends on depth. P = P0 + rho g h will be given to you. What will not be given is the reason the shape of the vessel and the amount of liquid do not appear in it, and that omission is where the marks are.

The second idea is Archimedes' principle: the upthrust on an object equals the weight of the fluid it displaces. Every buoyancy question is that one sentence plus a decision about which volume 'displaced' refers to. A floating object displaces its own weight; a fully submerged one displaces its own volume. Settle which case you are in before you touch a number, because that decision is the question.

What to hold

• Pressure in a liquid at rest rises with vertical depth as P = P0 + rho g h, and h is the vertical drop below the surface, not a distance along a slope or round a pipe.
• Pressure at a given depth does not depend on the shape of the vessel or on how much liquid it holds. The area cancels out of the column argument, and where the vessel is not a straight column its walls carry the difference.
• In one connected body of the same liquid at rest, any two points at the same level are at the same pressure. That single rule is the whole of manometer reading.
• Gauge pressure is the amount by which pressure exceeds atmospheric. A manometer open to the air reads gauge pressure, and the height difference between the two surfaces is what carries the answer.
• Pascal's principle says pressure applied to an enclosed fluid is transmitted undiminished, so a hydraulic system multiplies force by the ratio of the piston areas. It does not multiply work: the large piston moves proportionally less far.
• Archimedes' principle: the upthrust equals the weight of fluid displaced, which is rho(fluid) x V(displaced) x g.
• A floating object displaces its own weight of fluid, because the upthrust balances gravity. A fully submerged object displaces its own volume, whatever it weighs.
• For a floating object the fraction of its volume below the surface equals the ratio of its density to the fluid's, so anything denser than the fluid cannot float in it.
• The upthrust on a fully submerged object does not change with depth. It comes from the pressure difference between the top and bottom faces, which is set by the object's own height, not by how far down it has been taken.
• Apparent weight is true weight minus upthrust, so an object hanging submerged from a balance reads light by exactly the weight of fluid it displaces.
• Upthrust depends on the fluid's density and the displaced volume. It never depends on the object's density or on what the object is made of.