Fluid Pressure and Buoyancy
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Direct answer
Pressure at depth h in a liquid of density rho is P = P(0) + rho g h, acting equally in all directions — Pascal's law generalises this: pressure applied to an enclosed fluid transmits undiminished throughout, so a hydraulic lift multiplies force by the area ratio F(2) = F(1) A(2)/A(1). Archimedes' principle states that a body immersed in a fluid loses weight equal to the weight of fluid displaced; the submerged fraction while floating equals the density ratio — ice floating in water keeps exactly 90% below the surface. Standard anchors: 1 atmosphere = 1.013 × 10^5 Pa = 76 cm of mercury, and gauge pressure is P(absolute) minus atmospheric.
What you must remember
- Depth relation: P = P(0) + rho g h; pressure depends on depth, not on the container's shape — the hydrostatic paradox that decides many assertion–reason questions.
- Pascal's law: enclosed fluids transmit pressure equally in all directions; hydraulic machines are force multipliers, not energy multipliers — the small piston travels farther in the same ratio.
- Buoyant force: F(B) = rho(fluid) × V(displaced) × g, always upward through the centre of buoyancy; it equals the weight of displaced fluid whether the body floats or sinks.
- Floating fraction: submerged fraction = density of body/density of liquid; ice (900 kg/m^3) floats 90% immersed in water, and melting floating ice leaves the water level unchanged.
- Apparent weight: a body weighing W in air weighs W − F(B) when immersed; a spring balance plus a beaker on a weighing scale swap exactly the buoyant force between them.
- Barometer: 76 cm of mercury = 1.013 × 10^5 Pa at sea level; mercury is chosen because its density (13.6 × 10^3 kg/m^3) keeps the column manageable and its vapour pressure is negligible.
- Gauge versus absolute: gauge pressure = absolute − atmospheric; tyre gauges and most manometers read gauge.
- Pattern note: Main asks direct P = rho g h and floatation-fraction numericals; Advanced dresses buoyancy in accelerated-lift and two-liquid problems.
Reading a buoyancy problem honestly
A block of density 600 kg/m^3 floats at the interface of oil (800) above water (1000). Let V be the block's volume and x the fraction in water: the buoyant force is rho(oil) g V(1 − x) + rho(water) g V x, and floating makes it equal the block's weight 600 V g. Dividing out: 800(1 − x) + 1000x = 600 — the oil contributes more than the block's whole weight, so the block rides above the interface entirely in the oil, with 600/800 = 75% submerged there and 25% in air.
The hydrometer inverts the same logic. Its stem is thin so that small density changes produce large visible excursions: floating in liquid of density rho, the submerged volume is m/rho, and calibrating the stem in density units makes it a direct-reading instrument — the reason lactometers (milk, density about 1030 kg/m^3) carry a bulky bulb and a fine stem near the mark for adulterated samples.
Where students slip
The commonest fault is using the body's density inside the buoyant force; the displaced fluid's density is what counts — a stone's buoyancy in mercury differs from its buoyancy in water by the fluid alone. Second, at an interface or with a second fluid above, the weighted-sum balance of the depth paragraph is mandatory; the simple density-ratio fraction holds only for a single fluid. Third, the hydraulic lift violates no energy conservation: force is multiplied by A(2)/A(1) exactly as the piston displacement is divided by the same factor, so work in equals work out (ideally). Finally, in the ice-melting question — level unchanged for floating ice, level rising for an ice cube holding a stone that sinks when released — the reasoning, not a memorised verdict, is what Advanced marks.
Frequently asked questions
Does pressure at a point in a liquid depend on the shape of the container?
No — P = P(0) + rho g h depends only on depth, density and gravity; a narrow tube and a wide tank filled to the same height exert the same bottom pressure.
What does Pascal's law say and what device uses it?
Pressure applied to an enclosed fluid is transmitted undiminished to every point; the hydraulic lift and hydraulic brakes use it to multiply force by the area ratio.
What fraction of a floating body stays above the liquid?
The submerged fraction equals the ratio of body density to liquid density, so the emerged fraction is 1 − rho(body)/rho(liquid); ice shows 10% above water.
Why does the water level not change when floating ice melts?
The ice displaces exactly its own weight of water while floating, which is precisely the volume the meltwater occupies, so the level holds.
What does a body weigh when fully immersed?
Its weight in air minus the buoyant force rho(fluid) V g, read on a spring balance.