Upthrust in Fluids Archimedes Principle and Floatation
ICSE Class 9 · Physics · Chapter 5
What this chapter covers
- Buoyancy and upthrust
- Archimedes' principle
- A body in a liquid: the three cases
- Relative density
- Floatation
- Applications of floatation
- Numerical problems on all three cases
Key points
- Upthrust (buoyant force) is the upward force of a fluid on an immersed body; it arises because pressure increases with depth.
- Upthrust FB = Vρ g depends on the immersed volume and the density of the fluid, not on the body's material or weight.
- Archimedes' principle: upthrust equals the weight of fluid displaced.
- Apparent weight = true weight − upthrust; loss of weight = upthrust.
- Case I (ρB > ρL): sinks; Case II (ρB = ρL): floats fully immersed at any depth; Case III (ρB < ρL): rises and floats partly immersed. A floating body has zero apparent weight.
- Relative density = density of substance ÷ density of water at 4 °C; it has no unit.
- RD of a solid = weight in air ÷ loss of weight in water; RD of a liquid = loss in liquid ÷ loss in water.
- Principle of floatation: weight of floating body = weight of liquid displaced.
- Fraction immersed: ρ1 / ρ2 = V2 / V1.
- Ships, submarines, icebergs, balloons and life jackets are all explained by floatation.
Questions answered
Define upthrust and state its SI unit.
Upthrust is the upward force exerted by a fluid on a body wholly or partly immersed in it. Its SI unit is the newton.
Explain why upthrust arises, in terms of pressure.
Pressure increases with depth, so the upward thrust on the lower surface of an immersed body is greater than the downward thrust on its upper surface; the difference is the upthrust.
Name two factors on which upthrust depends, and one on which it does not.
It depends on the volume of the body immersed and the density of the fluid; it does not depend on the material or weight of the body.
State Archimedes' principle.
A body wholly or partially immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced by it.
What is meant by apparent weight? Why does a body appear to lose weight in water?
Apparent weight is the weight of a body measured while it is immersed in a fluid, equal to its true weight minus the upthrust. The upthrust supports part of the weight, so the balance reads less.
When does a body sink in a liquid, and when does it float fully immersed at any depth?
It sinks when its density is greater than that of the liquid (FB < W); it stays at any depth when the densities are equal (FB = W).
Define relative density. Why has it no unit?
Relative density is the ratio of the density of a substance to the density of water at 4 °C. It is a ratio of two densities, so the units cancel.
The relative density of silver is 10.5. What is its density in SI units?
10.5 × 1000 = 10 500 kg m-3.
Past board exam questions
ICSE 2019 sample paper · 2 marks
A body is completely immersed in a fluid. State two factors on which the upthrust acting on the body depends.
Show answer
1. The volume of the body immersed in the fluid (here the whole volume of the body).
2. The density of the fluid.
(Also the acceleration due to gravity: upthrust = V ρ g.)
ICSE 2019 sample paper · 3 marks
(i) Why does a piece of steel sink in water but float on mercury? (ii) If a bowl is formed from the same steel piece then it can float in water. Why?
Show answer
(i) The density of steel (about 7.8 g cm⁻³) is greater than that of water (1 g cm⁻³), so the upthrust of water on the piece is less than its weight and it sinks. The density of mercury (13.6 g cm⁻³) is greater than that of steel, so the piece floats on mercury when it is partly immersed and the upthrust equals its weight.
(ii) When made into a bowl, the steel encloses air, so its average density (mass ÷ total volume) becomes less than that of water. It displaces a much larger volume of water, so the upthrust becomes equal to its weight before it is fully immersed and the bowl floats.
ICSE 2019 sample paper · 4 marks
A metal piece weighs 200 gf in air and 150 gf when completely immersed in water. (i) Calculate the relative density of the metal piece. (ii) How much will it weigh in a liquid of density 0.8 gcm⁻³?
Show answer
(i) Loss of weight in water = 200 − 150 = 50 gf.
R.D. = weight in air ÷ loss of weight in water = 200 ÷ 50 = 4.
(ii) Volume of the piece = 50 cm³ (it displaces 50 g of water). Upthrust in the liquid = 50 × 0.8 = 40 gf.
Weight in the liquid = 200 − 40 = 160 gf.