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IMAT Physics · Formula sheet

Fluids formulas

5 formulas and 7 common traps for IMAT Physics Fluids, grouped by subtopic.

Full notes with worked examples

Pressure, Depth and the Hydraulic Press

Learn this subtopic in the notes

Pressure as force per area, and its units

Pressure

p=FAp = \frac{F}{A}
  • pppressure, in Pa
  • FFforce perpendicular to the surface, in N
  • AAarea, in m²

Pressure at a depth in a liquid (Stevin's law)

Pressure at depth h

p=p0+ρghp = p_0 + \rho g h
  • p0p_0pressure at the surface (often atmospheric)
  • ρ\rhodensity of the liquid, in kg/m³
  • gggravitational field strength, about 9.8 N/kg (IMAT often says 10)
  • hhdepth below the surface, in m

Pascal's principle and the hydraulic press

Hydraulic press

F1A1=F2A2A1d1=A2d2\frac{F_1}{A_1} = \frac{F_2}{A_2} \qquad A_1 d_1 = A_2 d_2
  • F1,A1,d1F_1, A_1, d_1force on, area of, and distance moved by the small piston
  • F2,A2,d2F_2, A_2, d_2the same for the large piston

Common traps

Square millimetres are not a thousandth of a square metre

A millimetre is 10−310^{-3} m, so a square millimetre is (10−3)2=10−6 m2(10^{-3})^2 = 10^{-6}\ \text{m}^2. Converting the length and forgetting to square it gives an answer a thousand times too small, and IMAT puts that wrong answer among the options.

Doubling the pressure means doubling the absolute pressure

At the surface the pressure is already about 1 atm from the air. It doubles when the water adds another 1 atm, at about 10 m. Answering with the depth where the water's own pressure doubles has no meaning, because that pressure starts at zero.

A still liquid pushes perpendicular to the wall

A liquid at rest cannot push along a surface; any sideways component would make it flow. So the force on each part of a wall is at right angles to that part, at every point. Options saying the force can be in any direction, or parallel to the wall, are wrong.

A hydraulic press multiplies force, not energy

The output force is larger by the area ratio, but the output piston moves a distance smaller by the same ratio. Work in equals work out (ignoring friction). An option claiming the press gives out more energy than it takes in is always wrong.

Upthrust, Floating and Sinking

Learn this subtopic in the notes

Archimedes' principle: the size of the upthrust

Upthrust

FB=ρfluid Vsub gF_B = \rho_{\text{fluid}}\, V_{\text{sub}}\, g
  • ρfluid\rho_{\text{fluid}}density of the fluid, in kg/m³
  • VsubV_{\text{sub}}volume of the body below the surface, in m³
  • gggravitational field strength, in N/kg

Floating and sinking by comparing densities

Fraction submerged when floating

VsubV=ρobjectρliquid\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}}
  • VsubV_{\text{sub}}volume below the surface
  • VVtotal volume of the object

Common traps

The upthrust uses the liquid's density

The upthrust on a fully submerged body is ρfluidVg\rho_{\text{fluid}} V g. Writing the object's own density there gives its weight, not the upthrust. An option of the form (ρfluid−ρobject)Vg(\rho_{\text{fluid}} - \rho_{\text{object}}) V g is the net force, not the upthrust.

A sunken object still feels upthrust

An object lying on the bottom is still surrounded by liquid, so the upthrust still acts on it. The bottom pushes up only with the weight minus the upthrust. Answering with the full weight leaves the upthrust out.

For a floating object, the upthrust equals the weight in every liquid

Move a floating block from water to oil and the upthrust does not change; it still equals the weight. What changes is how much of the block is under the surface. Options saying the upthrust is smaller in the less dense liquid are wrong while the block still floats.

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