PhysicsICSE Class 9 · Upthrust, Archimedes' Principle and Floatation

Floatation and Relative Density: Fraction Immersed, Ships and Submarines (ICSE Class 9)

Q: What is relative density?

The ratio of the density of a substance to the density of water at 4 °C. It has no unit.

Q: What fraction of a floating body is under water?

The ratio of the body’s density to the liquid’s density.

Q: Why does a steel ship float but a steel needle sink?

The ship’s hollow shape displaces a volume of water whose weight equals the ship’s weight, so its average density is below that of water.

Updated 17 October 2026 · Fact-checked

A floating body displaces fluid whose weight equals its own weight, so the fraction of its volume immersed is ρ_body/ρ_fluid. Relative density = density of substance ÷ density of water (no unit); for a solid heavier than water, RD = W₁/(W₁ − W₂) and for a liquid RD = (W₁ − W₃)/(W₁ − W₂).

Weight equals upthrust in both tanks, but the denser liquid needs less volume immersed: fraction = block density ÷ liquid density.

Illustration drawn for QuestionBuddy.

Understand Floatation, Relative Density and Fraction Immersed

Principle of floatation. A floating body displaces a weight of fluid equal to its own weight. Equivalently, the upthrust equals the weight, and its apparent weight is zero.

For a body of volume $V$ and density $ρ$ , floating with volume $V^{'}$ immersed in a liquid of density $ρ_{L}$ :

$V ρ g = V^{'} ρ_{L} g \Rightarrow \frac{V ^{'}}{V} = \frac{ρ}{ρ _{L}} .$

The fraction immersed equals the ratio of the densities.
A body floats only if $ρ < ρ_{L}$ (fraction less than 1). If the ratio is exactly 1 it stays wherever placed; more than 1 and it sinks.
In a denser liquid less of the body is submerged: a ship floats higher in sea water than in river water, but displaces the same weight of water in each.

Added load. A floating body with an extra load sinks until the extra upthrust equals the extra weight. The extra volume immersed is $\frac{extra mass}{ρ _{L}}$ . A body just sinks fully when the load brings the total weight up to the upthrust at full immersion, $V ρ_{L} g$ .

Relative density (RD). $RD = \frac{density of the substance}{density of water at 4 °C}$ ; also called specific gravity. It has no unit. Density in kg m⁻³ is $RD \times 1000$ . A body floats in water if its RD is less than 1.

RD of a solid heavier than water. Weigh it in air ( $W_{1}$ ) and fully in water ( $W_{2}$ ). The loss $W_{1} - W_{2}$ is the weight of an equal volume of water, so
$RD = \frac{W _{1}}{W _{1} - W _{2}} .$

RD of a liquid. Use a solid that sinks in both water and the liquid. Weights: in air $W_{1}$ , in water $W_{2}$ , in the liquid $W_{3}$ . The losses are the weights of equal volumes of water and liquid:
$RD_{l i q u i d} = \frac{W _{1} - W _{3}}{W _{1} - W _{2}} .$

Applications.

Ship: hollow steel hull displaces a large volume of water, so the upthrust equals its weight.
Submarine: ballast tanks filled with water raise its weight until it sinks or hovers; compressed air blows the water out so that it rises.
Hydrometer: a weighted tube floats deeper in a lighter liquid. For a uniform tube $\frac{h _{1}}{h _{2}} = \frac{ρ _{2}}{ρ _{1}}$ (lengths submerged in two liquids).
Hot-air or helium balloon: rises because the upthrust of the air displaced exceeds its total weight.

Floatation, Relative Density and Fraction Immersed: formulas and key facts

Floatation: $V ρ = V^{'} ρ_{L} \Rightarrow \frac{V ^{'}}{V} = \frac{ρ}{ρ _{L}}$; Fraction immersed.
RD of a solid: $RD = \frac{W _{1}}{W _{1} - W _{2}}$; Heavier than water.
RD of a liquid: $RD = \frac{W _{1} - W _{3}}{W _{1} - W _{2}}$; Same solid in air, water and the liquid.
Hydrometer-type tube: $h_{1} ρ_{1} = h_{2} ρ_{2}$; Same tube, same weight.

How to solve Floatation, Relative Density and Fraction Immersed questions

For floating bodies equate weight and upthrust; for RD use losses of weight.

1For floating: weight of body = weight of liquid displaced.
2Write the immersed fraction as $ρ / ρ_{L}$ and solve for the unknown.
3For a load: extra mass = extra volume immersed × $ρ_{L}$ .
4For RD: compute the losses of weight, then take the ratio.
5Remember RD has no unit; density = RD × 1000 kg m⁻³.

Floatation, Relative Density and Fraction Immersed 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.

A loaded boat

Suppose a flat boat with a waterline area of 8 m² in fresh water (1000 kg m⁻³) takes on 400 kg of cargo.

How much deeper does it sink?

Try it first, then show the solutionSolution

Extra volume displaced $= \frac{400}{1000} = 0.4$ m³.
Extra depth $= \frac{0.4}{8} = 0.05$ m.

Answer: 5 cm deeper.

Memory hook: Extra mass ÷ ρ_L gives the extra volume.

Tricks to remember Floatation, Relative Density and Fraction Immersed

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

Fraction immersed = density ratio
Body density ÷ liquid density.
Use it when: Floating questions.
RD solid: air over loss
W₁ over (W₁ − W₂).
Use it when: Experiment answers.
RD liquid: loss over loss
Loss in liquid over loss in water.
Use it when: Three-weighing method.

Floatation, Relative Density and Fraction Immersed: solved examples

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

Example 1: Wood in water · 1 mark · MCQBoard level

A wooden cube floats in water with three-quarters of its volume immersed. The relative density of the wood is:
(a) 0.25
(b) 0.50
(c) 0.75
(d) 1.33

Try it first, then show the solutionSolution

$\frac{V ^{'}}{V} = \frac{ρ}{ρ _{w}} = 0.75$ .

Answer: (c) 0.75

Example 2: RD of a metal · 2 marks · Short answerBoard level

A piece of metal weighs 3.6 N in air and 3.2 N when fully immersed in water. Find its relative density and its density.

Try it first, then show the solutionSolution

Loss $= 3.6 - 3.2 = 0.4$ N. (½)
$RD = \frac{3.6}{0.4} = 9$ . (1)
Density $= 9 \times 1000 = 9000$ kg m⁻³. (½)

Answer: RD 9; 9000 kg m⁻³

Example 3: RD of a liquid · 3 marks · Short answerHarder board level

A solid weighs 150 gf in air, 100 gf in water and 70 gf in a liquid. Find the RD of the solid and of the liquid.

Try it first, then show the solutionSolution

Loss in water $= 50$ gf. (½)
$RD_{so l i d} = \frac{150}{50} = 3$ . (1)
Loss in liquid $= 150 - 70 = 80$ gf; $RD_{l i q u i d} = \frac{80}{50} = 1.6$ . (1½)

Answer: Solid 3; liquid 1.6

Example 4: Ship from sea to river · 3 marks · Short answerHarder board level

A ship of mass $4.12 \times 1 0^{6}$ kg floats in sea water of density 1030 kg m⁻³ and then enters a river (density 1000 kg m⁻³). Its waterline area is 800 m². Find the volume of water it displaces in each, and how much deeper it floats in the river.

Try it first, then show the solutionSolution

Sea: $V = \frac{4.12 \times 1 0 ^{6}}{1030} = 4000$ m³. (1)
River: $V = \frac{4.12 \times 1 0 ^{6}}{1000} = 4120$ m³. (1)
Extra $= 120$ m³; depth $= \frac{120}{800} = 0.15$ m. (1)

Answer: 4000 m³; 4120 m³; 0.15 m deeper

Example 5: A loaded test tube · 4 marks · Case-basedHOTS / competency

A uniform test tube loaded with lead shot floats upright. It sinks to a depth of 12 cm in water, 10 cm in liquid X and 15 cm in liquid Y. Its cross-sectional area is 2 cm².
(a) Find the relative density of X.
(b) Find the relative density of Y.
(c) Which liquid is denser, and what instrument works on this idea?
(d) Find the mass of the loaded tube.

Try it first, then show the solutionSolution

Same weight, so $h ρ$ is constant. (1)
(a) $ρ_{X} = \frac{12}{10} = 1.2$ times that of water. (1)
(b) $ρ_{Y} = \frac{12}{15} = 0.8$ times. (1)
(c) X is denser (tube sinks less). The idea is used in the hydrometer. (½)
(d) Volume displaced in water $= 2 \times 12 = 24$ cm³, mass $= 24$ g. (½)

Answer: (a) 1.2 (b) 0.8 (c) X; hydrometer (d) 24 g

Where students lose marks in Floatation, Relative Density and Fraction Immersed

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

Mistake: Unit given for relative density
RD is a ratio.
Wrong working
RD = 9 kg m⁻³
Right working
RD = 9
Do this: No unit for RD; density has a unit.
Usually costs: The unit mark.
Mistake: Fraction inverted
Fraction immersed = body density ÷ liquid density.
Wrong working
V'/V = ρ_L/ρ
Right working
V'/V = ρ/ρ_L
Do this: Check the fraction is less than 1 for a floating body.
Usually costs: The answer.
Mistake: Losses mixed up in the liquid method
Liquid RD = loss in liquid ÷ loss in water.
Wrong working
RD = W₃/W₂
Right working
RD = (W₁ − W₃)/(W₁ − W₂)
Do this: Compute both losses explicitly.
Usually costs: The answer.

Exam tips

State the principle of floatation, then write weight = upthrust.
Show the fraction immersed as ρ/ρ_L and check it is less than 1.
Give RD without a unit, and density in kg m⁻³ with one.
In ship questions say the weight of water displaced is the same in both waters; only the volume changes.

Practice questions on Floatation, Relative Density and Fraction Immersed

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
A floating body displaces a weight of liquid that is:
(a) less than its own weight
(b) equal to its own weight
(c) greater than its own weight
(d) zero
Hint
Principle of floatation.
Answer
(b)
Q2 · 1 mark · Very short answerBoard level
Does relative density have a unit?
Hint
It is a ratio.
Answer
No.
Q3 · 2 marks · Short answerHarder board level
A body floats in a liquid of density 1250 kg m⁻³ with 0.64 of its volume immersed. Find its density.
Hint
Fraction = ρ/ρ_L.
Answer
$ρ = 0.64 \times 1250 = 800$ kg m⁻³.
Q4 · 3 marks · Short answerHarder board level
A solid of mass 2 kg and volume 2500 cm³ is placed in water. (a) Does it float? (b) What fraction is immersed? (c) What extra mass placed on it will just make it fully immersed?
Hint
Density = 800 kg m⁻³.
Answer
(a) Density $= \frac{2}{0.0025} = 800$ kg m⁻³ < 1000: floats. (b) 0.8. (c) Water of the full volume has mass 2.5 kg, so the extra mass is $2.5 - 2 = 0.5$ kg.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A ship floats higher in sea water than in river water.
Reason (R): Sea water is denser, so a smaller volume of it has to be displaced to balance the ship’s weight.
(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
Weight fixed.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A submarine has a hull volume of 5000 m³ and a mass of $4.8 \times 1 0^{6}$ kg when its ballast tanks are empty. How much sea water (density 1000 kg m⁻³) must be let into the ballast tanks so that it just hovers fully submerged? Why must it blow water out to rise?
Hint
Hovering: weight = upthrust at full volume.
Answer
Upthrust $= 5000 \times 1000 \times g$ corresponds to a mass of $5 \times 1 0^{6}$ kg. Water to add $= 5 \times 1 0^{6} - 4.8 \times 1 0^{6} = 2 \times 1 0^{5}$ kg. To rise, the weight must fall below the upthrust, so water is blown out.

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.Principle of floatation?Show answer
A floating body displaces fluid of weight equal to its own weight.
2.Fraction immersed?Show answer
ρ_body/ρ_liquid.
3.RD of a solid heavier than water?Show answer
W₁/(W₁ − W₂).
4.RD of a liquid?Show answer
(W₁ − W₃)/(W₁ − W₂).
5.How does a submarine rise?Show answer
Compressed air blows water out of its ballast tanks, reducing its weight below the upthrust.

Floatation, Relative Density and Fraction Immersed: frequently asked questions

What is relative density?

The ratio of the density of a substance to the density of water at 4 °C. It has no unit.

What fraction of a floating body is under water?

The ratio of the body’s density to the liquid’s density.

Why does a steel ship float but a steel needle sink?

The ship’s hollow shape displaces a volume of water whose weight equals the ship’s weight, so its average density is below that of water.