PhysicsICSE Class 9 · Upthrust, Archimedes' Principle and Floatation

Archimedes' Principle: Upthrust, Apparent Weight and Loss of Weight (ICSE Class 9)

Q: What is Archimedes' principle?

When a body is wholly or partly immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced by it.

Q: Why does a body weigh less in water?

The upthrust acts upward and reduces the force that the supporting spring balance has to provide; apparent weight = W − F_B.

Q: Does the upthrust depend on the material of the body?

No, only on the volume immersed and the density of the fluid.

Updated 17 October 2026 · Fact-checked

Archimedes' principle: a body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid it displaces, F = Vρ_L g (V the immersed volume, ρ_L the fluid density). Apparent weight = true weight − upthrust, so the loss of weight in a liquid equals the upthrust.

The block loses 2 N of weight in water, and the water it pushes out into the beaker also weighs 2 N.

Illustration drawn for QuestionBuddy.

Understand Archimedes' Principle, Upthrust and Apparent Weight

Upthrust. A fluid pushes a body immersed in it upward. This force is the upthrust (buoyant force). It arises because pressure increases with depth, so the fluid pushes up on the bottom of the body harder than it pushes down on the top.

Archimedes’ principle. When a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.

$F_{B} = V ρ_{L} g$

$V$ = volume of the immersed part (equal to the volume of fluid displaced);
$ρ_{L}$ = density of the fluid (not of the body);
the upthrust acts vertically upward through the centre of gravity of the displaced fluid.

What it depends on, and what it does not. Upthrust increases with the immersed volume and with the fluid’s density. For a fully immersed body it does not depend on the depth, the shape, or the material or mass of the body. Two cubes of equal volume, one of iron and one of wood, held fully under water feel the same upthrust.

Apparent weight and loss of weight.

$apparent weight = W - F_{B}, loss of weight = F_{B} .$

That is why a stone feels lighter in water. The loss is greater in a denser liquid (sea water) than in a lighter one (kerosene).

Verification. Weigh a body in air ( $W_{1}$ ) and then fully in water ( $W_{2}$ ) on a spring balance. Collect the water pushed out of a full eureka (overflow) can and weigh it. The weight of the water collected equals the loss of weight $W_{1} - W_{2}$ .

Newton’s third law and the balance. The fluid pushes up on the body with $F_{B}$ , so the body pushes down on the fluid with $F_{B}$ . A beaker of water on a scale reads more by $F_{B} / g$ in mass when a body is held in it, even if the body hangs from a string and does not touch the beaker.

Unit care. In gf: weight of 1 cm³ of water is 1 gf, so the upthrust on $V$ cm³ in water is $V$ gf. In N: use $V$ in m³, $ρ$ in kg m⁻³, $g = 10$ or $9.8$ .

Three cases for a fully immersed body. $ρ_{b o d y} > ρ_{L}$ : sinks; $ρ_{b o d y} = ρ_{L}$ : stays anywhere; $ρ_{b o d y} < ρ_{L}$ : rises and floats (see the floatation topic).

Archimedes' Principle, Upthrust and Apparent Weight: formulas and key facts

Upthrust: $F_{B} = V ρ_{L} g$; V = immersed volume; ρ_L = fluid density.
Apparent weight: $W_{a pp} = W - F_{B}$; Loss of weight = upthrust.
Volume from loss: $V = \frac{W _{1} - W _{2}}{ρ _{L} g}$; Weights in N, V in m³.

How to solve Archimedes' Principle, Upthrust and Apparent Weight questions

Find the immersed volume and the fluid density first.

1Decide how much of the body is below the surface (V).
2Pick the correct density: that of the fluid, not of the body.
3Compute $F_{B} = V ρ_{L} g$ (or loss of weight if weights are given).
4Apparent weight = true weight − upthrust.
5Use the loss of weight to recover V and then the body’s density if asked.

Archimedes' Principle, Upthrust and Apparent Weight in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Lifting a boulder in a stream

Suppose a boulder of volume 0.04 m³ and density 2500 kg m⁻³ sits at the bottom of a stream.

Compare the force needed to lift it in air with the force needed to lift it while it is under water (g = 10 m s⁻²).

Try it first, then show the solutionSolution

Weight $= 0.04 \times 2500 \times 10 = 1000$ N.
Upthrust $= 0.04 \times 1000 \times 10 = 400$ N, so the pull needed under water is $600$ N.

Answer: 1000 N in air and 600 N under water.

Memory hook: Upthrust does the 400 N of lifting for you.

Tricks to remember Archimedes' Principle, Upthrust and Apparent Weight

Short memory aids. Say each one aloud once, then use it on a question the same day.

Weight of fluid displaced
Upthrust = weight of the water that got pushed out of the way.
Use it when: Statement and numericals.
Fluid’s density, not the body’s
In Vρg, ρ belongs to the fluid.
Use it when: Choosing the density.
Loss of weight = upthrust
W in air − W in liquid = F_B.
Use it when: Experiment-based questions.

Archimedes' Principle, Upthrust and Apparent Weight: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Apparent weight · 1 mark · MCQBoard level

A body weighs 30 N in air and experiences an upthrust of 12 N in a liquid. Its apparent weight in the liquid is:
(a) 42 N
(b) 30 N
(c) 18 N
(d) 12 N

Try it first, then show the solutionSolution

$30 - 12 = 18$ N.

Answer: (c) 18 N

Example 2: Upthrust in two liquids · 2 marks · Short answerBoard level

A stone of volume 400 cm³ is fully immersed in (a) water and (b) kerosene of density 800 kg m⁻³. Find the upthrust in each ( $g = 10$ m s⁻²).

Try it first, then show the solutionSolution

$V = 400 \times 1 0^{- 6} = 4 \times 1 0^{- 4}$ m³. (½)
(a) $4 \times 1 0^{- 4} \times 1000 \times 10 = 4$ N. (1)
(b) $4 \times 1 0^{- 4} \times 800 \times 10 = 3.2$ N. (½)

Answer: (a) 4 N (b) 3.2 N

Example 3: Overflow can · 3 marks · Short answerHarder board level

A body of mass 0.5 kg and volume 200 cm³ is lowered into a eureka can filled to the spout. It sinks. Find the mass and weight of water that overflows, the upthrust and the apparent weight in water ( $g = 10$ m s⁻²).

Try it first, then show the solutionSolution

Water displaced $= 200$ cm³ $= 0.2$ kg of water, weight $0.2 \times 10 = 2$ N. (1)
Upthrust $= 2$ N. (1)
Weight of body $= 5$ N; apparent weight $= 5 - 2 = 3$ N. (1)

Answer: 0.2 kg (2 N); upthrust 2 N; apparent weight 3 N

Example 4: Density from loss of weight · 3 marks · Short answerHarder board level

A metal sphere weighs 7.2 N in air and 6.3 N when fully immersed in water ( $g = 10$ m s⁻², $ρ_{w} = 1000$ kg m⁻³). Find (a) its volume, (b) its density, (c) its apparent weight in a liquid of density 800 kg m⁻³.

Try it first, then show the solutionSolution

(a) Loss $= 0.9$ N $= V \times 1000 \times 10$ , so $V = 9 \times 1 0^{- 5}$ m³. (1)
(b) Mass $= 0.72$ kg; $ρ = \frac{0.72}{9 \times 1 0 ^{- 5}} = 8000$ kg m⁻³. (1)
(c) Upthrust $= 9 \times 1 0^{- 5} \times 800 \times 10 = 0.72$ N; apparent weight $= 7.2 - 0.72 = 6.48$ N. (1)

Answer: (a) 9 × 10⁻⁵ m³ (b) 8000 kg m⁻³ (c) 6.48 N

Example 5: A beaker on a scale · 4 marks · Case-basedHOTS / competency

A beaker of water stands on a scale that reads 2.000 kg. A stone of volume 300 cm³ and density 2500 kg m⁻³ is hung from a string and lowered fully into the water without touching the beaker. ( $g = 10$ m s⁻², $ρ_{w} = 1000$ kg m⁻³.)
(a) Find the upthrust on the stone.
(b) What does the scale now read?
(c) What force does the stone exert on the water? Name the law.
(d) The string is released and the stone rests on the bottom. What does the scale read now?

Try it first, then show the solutionSolution

(a) $300 \times 1 0^{- 6} \times 1000 \times 10 = 3$ N. (1)
(b) By the third law the water gets pushed down with 3 N: the scale reads $2.000 + 0.300 = 2.300$ kg. (1)
(c) 3 N downward on the water; Newton’s third law. (1)
(d) The whole stone is now supported by the beaker: stone mass $= 300 \times 2.5 = 750$ g, so the scale reads $2.000 + 0.750 = 2.750$ kg. (1)

Answer: (a) 3 N (b) 2.300 kg (c) 3 N (d) 2.750 kg

Where students lose marks in Archimedes' Principle, Upthrust and Apparent Weight

The slips examiners see most often, and what to write instead.

Mistake: Body’s density used in Vρg
ρ is the fluid’s density.
Wrong working
F = V × 8000 × 10
Right working
F = V × 1000 × 10 (water)
Do this: Label ρ_L explicitly.
Usually costs: The answer.
Mistake: Cm³ not converted to m³
SI units needed.
Wrong working
F = 400 × 1000 × 10
Right working
F = 4 × 10⁻⁴ × 1000 × 10 = 4 N
Do this: Convert V first.
Usually costs: The answer.
Mistake: Upthrust said to depend on depth
Fully immersed: independent of depth.
Wrong working
The deeper the body, the greater the upthrust.
Right working
It stays the same once fully immersed.
Do this: Remember it depends on volume and fluid density only.
Usually costs: The reasoning mark.

Exam tips

State Archimedes’ principle in full, including "wholly or partly immersed".
Write which density you are using in F = Vρg.
Apparent weight must be less than true weight; check this.
For a floating body, upthrust equals the weight of the body.

Practice questions on Archimedes' Principle, Upthrust and Apparent Weight

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The upthrust on a fully immersed body acts:
(a) vertically downward
(b) vertically upward through the centre of gravity of the displaced fluid
(c) horizontally
(d) along the surface
Hint
Direction of buoyancy.
Answer
(b)
Q2 · 1 mark · Very short answerBoard level
What is a eureka can used for?
Hint
Overflow.
Answer
To collect the liquid displaced by a body and so find its volume or the weight of the liquid displaced.
Q3 · 2 marks · Short answerHarder board level
Find the upthrust on a fully immersed body of volume 0.05 m³ in sea water of density 1030 kg m⁻³ ( $g = 10$ m s⁻²).
Hint
F = Vρg.
Answer
$F = 0.05 \times 1030 \times 10 = 515$ N.
Q4 · 3 marks · Short answerHarder board level
A block of mass 0.6 kg and volume 800 cm³ is held fully under water and then released ( $g = 10$ m s⁻²). Find the net force on it at the moment of release and say what happens.
Hint
Compare weight and upthrust.
Answer
Weight $= 6$ N; upthrust $= 8 \times 1 0^{- 4} \times 1000 \times 10 = 8$ N. Net force $= 2$ N upward: the block rises and floats.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): The upthrust on a fully immersed body does not change as it is lowered deeper.
Reason (R): Upthrust depends only on the volume of the body and the density of the fluid.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Vρg.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A solid appears to weigh three-quarters of its true weight when fully immersed in water. Find its relative density and its apparent weight, as a fraction of its true weight, in a liquid of density 800 kg m⁻³.
Hint
Loss = one-quarter of W.
Answer
Loss in water $= \frac{W}{4}$ , so the RD is $\frac{W}{W /4} = 4$ . Upthrust scales with the liquid’s density: loss in the liquid $= \frac{W}{4} \times 0.8 = 0.2 W$ , so apparent weight $= 0.8 W$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for upthrust?Show answer
F_B = Vρ_L g.
2.Loss of weight in a liquid equals?Show answer
The upthrust.
3.Direction of upthrust?Show answer
Vertically upward.
4.Why does a fully immersed body feel the same upthrust at any depth?Show answer
Upthrust depends only on the volume immersed and the density of the fluid.
5.What is apparent weight?Show answer
True weight minus upthrust.

Archimedes' Principle, Upthrust and Apparent Weight: frequently asked questions

What is Archimedes' principle?

When a body is wholly or partly immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced by it.

Why does a body weigh less in water?

The upthrust acts upward and reduces the force that the supporting spring balance has to provide; apparent weight = W − F_B.

Does the upthrust depend on the material of the body?

No, only on the volume immersed and the density of the fluid.