NDA Physics · Fluid Mechanics and Properties of Matter

Buoyancy, Density and Flotation

Density (mass per unit volume) decides everything here: a body floats when its average density is less than the fluid's, and Archimedes' principle says the upward buoyant force equals the weight of the fluid the body displaces.

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Why this matters

Sixteen PYQs — the largest and hardest pool in the chapter, with five HARD problems. The recurring tests are: density and relative density, Archimedes' upthrust (= weight of displaced fluid), the float-or-sink rule (compare densities), apparent weight loss when submerged, combining densities by mixing equal volumes versus equal masses, and the stability of a floating body (centre of gravity below the metacentre). Build the density foundation first — almost every hard problem here is a density comparison in disguise.

Concept 1 of 6

Density and relative density

Intuition

Density tells you how tightly mass is packed into a volume. A kilogram of lead takes up far less space than a kilogram of cotton, so lead is denser. Relative density just compares a substance's density with water's — it is a pure number with no units.

Definition

Density is mass per unit volume: $\rho = m/V$ (SI unit kg/m³).

Water has density 1000 kg/m³ (= 1 g/cm³), greatest near 4 °C.
Relative density (specific gravity) $= \rho_{\text{substance}} / \rho_{\text{water}}$ — a unitless ratio. RD = 0.8 means the substance is 0.8 times as dense as water.
A substance with RD < 1 floats on water; RD > 1 sinks.

Density and relative density

\rho = \dfrac{m}{V}, \qquad \text{RD} = \dfrac{\rho_{\text{substance}}}{\rho_{\text{water}}}

\rhodensity (kg/m³)
mmass (kg)
Vvolume (m³)
\text{RD}relative density (no unit)

Worked example

A pumpkin weighs 7.5 N. When fully submerged it displaces 0.75 litre of water. Find its density (g = 10 m/s², water = 1000 kg/m³).

Mass = weight / g = $7.5 / 10 = 0.75\,\text{kg}$ .
Volume displaced = body volume = $0.75\,\text{L} = 0.75 \times 10^{-3}\,\text{m}^3$ .
$\rho = m/V = 0.75 / (0.75 \times 10^{-3}) = 1000\,\text{kg/m}^3$ .

Answer:1000 kg/m³ (the same as water — it would just barely float).

Practice this conceptself-check · 3 quick reps

Try it yourself

A metal block has mass 540 g and volume 200 cm³. Find its density and its relative density (water = 1 g/cm³).

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Density of water at 4 °C?
2.
Unit of relative density?
3.
A substance of RD 0.7 in water — floats or sinks?

From the bank · past-year question

Example 1Fluid Mechanics and Properties of MatterMODERATE

A pumpkin weighs 7.5 N. On submerging completely in water, 3/4 L of water gets displaced. g = 10 m/s2. The correct value of density of the pumpkin is

[Q92 · Sep · 2024]

Relative density has no unit

RD is a ratio of two densities, so the units cancel — it is a pure number. An option that quotes RD 'in kg/m³' is wrong. Also remember water peaks in density near 4 °C, not at 0 °C.

Drill 2 more on density and relative density

Concept 2 of 6

Combining densities — equal volumes vs equal masses

Intuition

Mix two liquids and the result's density depends on HOW you mix them. Equal VOLUMES gives the plain average of the two densities. Equal MASSES gives the harmonic-style average — always smaller — because the lighter liquid takes up more room.

Definition

Mixing two substances of densities $\rho_1$ and $\rho_2$ :

Equal volumes ( $V$ each): average density $= \dfrac{\rho_1 + \rho_2}{2}$ (the arithmetic mean).
Equal masses ( $m$ each): average density $= \dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}$ (the harmonic mean of the two).

Average density is always total mass divided by total volume.

Mixture density (equal masses)

\rho_{\text{eq.mass}} = \dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}

\rho_1, \rho_2densities of the two components
\rho_{\text{eq.mass}}density of an equal-mass mixture

Worked example

Two substances of densities rho1 and rho2 are mixed in equal volumes giving relative density 4, and in equal masses giving relative density 3. Find rho1 and rho2.

Equal volumes: $(\rho_1+\rho_2)/2 = 4 \Rightarrow \rho_1+\rho_2 = 8$ .
Equal masses: $2\rho_1\rho_2/(\rho_1+\rho_2) = 3 \Rightarrow 2\rho_1\rho_2/8 = 3 \Rightarrow \rho_1\rho_2 = 12$ .
Solve $\rho_1+\rho_2=8$ , $\rho_1\rho_2=12$ : the roots are 6 and 2.

Answer:rho1 = 6 and rho2 = 2 (in units of water's density).

Practice this conceptself-check · 3 quick reps

Try it yourself

An object is built from two equal VOLUMES, one of density rho0 and the other of density 2 rho0. Find its average density.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Equal-volume mix of densities 2 and 6 — average density?
2.
Average density formula in general?
3.
Equal-mass mixing gives which kind of mean?

From the bank · past-year question

Example 2Fluid Mechanics and Properties of MatterHARD

Two substances of densities

\rho_1

and

\rho_2

are mixed in equal volume and their relative density is 4. When they are mixed in equal masses, relative density is 3. The values of

\rho_1

and

\rho_2

respectively are

[Q65 · Sep · 2019]

Equal volumes vs equal masses give DIFFERENT averages

Equal volumes means the simple average (rho1+rho2)/2. Equal masses means 2 rho1 rho2 / (rho1+rho2), which is smaller. Read the question carefully — using the wrong one is the classic mistake on this HARD favourite.

Drill 1 more on combining densities — equal volumes vs equal masses

Concept 3 of 6

Archimedes' principle and the buoyant force

Intuition

Lower an object into water and the water pushes up on it. Archimedes saw that this upthrust equals the weight of the water the object shoves aside. Displace more fluid and you get more lift — that is the whole principle.

Definition

Buoyancy (upthrust) is the upward force a fluid exerts on a body immersed in it. Archimedes' principle: the buoyant force equals the weight of the fluid displaced by the body: $F_b = \rho_{\text{fluid}}\, V_{\text{disp}}\, g$ .

It acts upward, through the centre of the displaced fluid (the centre of buoyancy).
It equals the weight of displaced fluid — not the mass or weight of the body itself.

Buoyant force (Archimedes)

F_b = \rho_{\text{fluid}}\, V_{\text{disp}}\, g

F_bbuoyant force / upthrust (N)
\rho_{\text{fluid}}density of the fluid (kg/m³)
V_{\text{disp}}volume of fluid displaced (m³)
gacceleration due to gravity (m/s²)

A floating body sinks until the weight of water it displaces equals its own weight. The submerged fraction equals the density ratio rho_body / rho_water.

Worked example

A stone of volume 200 cm³ is fully submerged in water (1000 kg/m³, g = 10 m/s²). Find the buoyant force on it.

Volume displaced = 200 cm³ = $200 \times 10^{-6}\,\text{m}^3$ .
$F_b = \rho V_{\text{disp}} g = 1000 \times 200 \times 10^{-6} \times 10$ .
$F_b = 2\,\text{N}$ — independent of the stone's own weight.

Answer:2 N upward.

Practice this conceptself-check · 3 quick reps

Try it yourself

When a body is immersed in a fluid the upthrust equals (1) the mass of the body, or (2) the weight of the fluid displaced by the body. Which is correct?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Buoyant force equals the weight of the ___?
2.
Direction of the buoyant force?
3.
Submerge the SAME body deeper (fully immersed already) — does buoyant force change?

From the bank · past-year question

Example 3Fluid Mechanics and Properties of MatterEASY

When a solid body is partially or completely immersed in a fluid, the fluid exerts an upward force on the body. The magnitude of the force is equal to 1. the mass of the body 2. the weight of the displaced fluid by the body Which of the above is/are correct?

[Q137 · Apr · 2025]

Upthrust = weight of displaced FLUID, not of the body

The buoyant force depends on the fluid's density and the displaced volume — never on the body's own mass or weight. An option claiming the upthrust equals 'the mass of the body' is wrong on two counts: it is the fluid's weight, and a mass is not a force.

Drill 2 more on archimedes' principle and the buoyant force

Concept 4 of 6

Apparent weight loss on submersion

Intuition

Lift a rock under water and it feels lighter — because the water pushes up on it. The scale reading drops by exactly the buoyant force, which is the weight of the water the rock displaces. The rock has not lost any mass; it has gained an upward helper.

Definition

A body submerged in a fluid weighs less on a scale because of the upthrust: Apparent weight = true weight − buoyant force = $W - \rho_{\text{fluid}} V_{\text{disp}} g$ .

The 'loss of weight' equals the buoyant force = weight of displaced fluid.
The body's true weight (and mass) is unchanged — only the SCALE reading falls.

Apparent weight in a fluid

W_{\text{app}} = W - F_b

W_{\text{app}}apparent (scale) weight in the fluid (N)
Wtrue weight in air (N)
F_bbuoyant force = weight of displaced fluid (N)

Worked example

A metal block weighs 50 N in air. Fully submerged in water it experiences a buoyant force of 8 N. What is its apparent weight in water?

Apparent weight = true weight − buoyant force.
$W_{\text{app}} = 50 - 8 = 42\,\text{N}$ .
The block FEELS lighter by 8 N (the weight of water it displaced).

Answer:42 N.

Practice this conceptself-check · 3 quick reps

Try it yourself

An object is weighed in air and then while fully submerged in water. How does its measured weight in water compare with its weight in air?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Why does a body weigh less under water?
2.
Loss in weight on submersion equals…
3.
Does the body's true mass change under water?

From the bank · past-year question

Example 4Fluid Mechanics and Properties of MatterEASY

If some object is weighed when submerged in water, what will happen to its weight compared to its weight in air?

[Q109 · Sep · 2017]

Mass is unchanged; only apparent weight drops

A submerged body does not lose mass or true weight — it only reads lighter on a scale because of the upthrust. The 'loss of weight' is exactly the buoyant force, equal to the weight of the displaced fluid.

Concept 5 of 6

Float or sink — the density comparison

Intuition

Whether something floats comes down to one comparison: average density of the body versus density of the fluid. Lighter-per-volume than the fluid, it floats; heavier, it sinks. A solid iron nail sinks, yet an iron ship floats — because the ship's hull traps air, dropping its AVERAGE density below water's.

Definition

Compare the body's average density $\rho_b$ with the fluid's density $\rho_f$ :

$\rho_b < \rho_f$ -> the body floats (part stays above the surface).
$\rho_b > \rho_f$ -> the body sinks.
$\rho_b = \rho_f$ -> it stays in neutral equilibrium (just submerged).

For a floating body, **fraction submerged $= \rho_b / \rho_f$ ** — and the weight of fluid displaced equals the body's full weight.

Fraction submerged of a floating body

\dfrac{V_{\text{submerged}}}{V_{\text{total}}} = \dfrac{\rho_{\text{body}}}{\rho_{\text{fluid}}}

V_{\text{submerged}}submerged volume (m³)
V_{\text{total}}total volume of the body (m³)
\rho_{\text{body}}average density of the body
\rho_{\text{fluid}}density of the fluid

Worked example

A sealed packet has volume 1 litre and mass 800 g. Will it float or sink in water (1 g/cm³) and in a liquid of density 1.5 g/cm³?

Packet density = $800\,\text{g} / 1000\,\text{cm}^3 = 0.8\,\text{g/cm}^3$ .
Water (1.0) and the liquid (1.5) are BOTH denser than 0.8.
A body floats whenever the fluid is denser than the body, so it floats in both.

Answer:It floats in water and in the 1.5 g/cm³ liquid (its 0.8 g/cm³ is less than both).

Practice this conceptself-check · 3 quick reps

Try it yourself

An iron nail sinks in water but an iron ship floats. Which statements are correct? (i) average density of the nail > water; (ii) average density of the ship < water.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
A body sinks when its density is ___ the fluid's.
2.
Float-or-sink depends on what difference?
3.
A block of RD 0.6 floats in water — what fraction is submerged?

From the bank · past-year question

Example 5Fluid Mechanics and Properties of MatterHARD

The volume of a sealed packet is 1 litre and its mass is 800 g. The packet is first put inside water with density 1 g cm

^{-3}

and then in another liquid

B

with density

1 \cdot 5

g cm

^{-3}

. Then which one of the following statements holds true?

[Q134 · Sep · 2022]

It is AVERAGE density that decides flotation

Iron sinks, yet an iron ship floats — because the hull traps air and lowers the SHIP's average density below water's. Never reason from the material alone; compare the body's average density with the fluid's.

Drill 5 more on float or sink — the density comparison

Concept 6 of 6

Stability of a floating body — the metacentre

Intuition

A floating ship can be tilted a little and still right itself — if its 'metacentre' sits above its centre of gravity. When the ship tilts, the centre of buoyancy shifts; the point where the new upthrust line crosses the ship's axis is the metacentre. Above the centre of gravity, the body is stable.

Definition

Three points govern floating stability:

Centre of gravity (G) — where the body's weight acts.
Centre of buoyancy (B) — the centre of the displaced fluid; the upthrust acts here.
Metacentre (M) — where the upthrust's line meets the body's axis after a small tilt.

Stable equilibrium requires the metacentre to lie ABOVE the centre of gravity (M above G). It does not require G below B — a tall ship can have G above B and still be stable, as long as M stays above G.

Worked example

A floating body is in stable equilibrium. What is the correct relationship between its centre of gravity and its metacentre?

Stability is decided by the metacentre M relative to the centre of gravity G.
For a restoring torque on a small tilt, M must lie ABOVE G.
So the centre of gravity is below the metacentre.

Answer:The centre of gravity lies below the metacentre (M above G).

Practice this conceptself-check · 3 quick reps

Try it yourself

A ship is loaded so that its centre of gravity rises above the metacentre after a small tilt. Is its equilibrium stable or unstable?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Condition for a floating body to be in stable equilibrium?
2.
At which point does the buoyant force act?
3.
If the metacentre is below the centre of gravity, the float is…

From the bank · past-year question

Example 6Fluid Mechanics and Properties of MatterMODERATE

A floating body is in stable state. Which one of the following is correct?

[Q59 · Apr · 2026]

Stability is about M above G, not G below B

The trap options compare the centre of gravity with the centre of BUOYANCY. Stability is actually decided by the METACENTRE: M must be above G. A floating body can have G above B and still be perfectly stable.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Density and relative density
Density and relative density
$\rho = \dfrac{m}{V}, \qquad \text{RD} = \dfrac{\rho_{\text{substance}}}{\rho_{\text{water}}}$
Combining densities — equal volumes vs equal masses
Mixture density (equal masses)
$\rho_{\text{eq.mass}} = \dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}$
Archimedes' principle and the buoyant force
Buoyant force (Archimedes)
$F_b = \rho_{\text{fluid}}\, V_{\text{disp}}\, g$
Apparent weight loss on submersion
Apparent weight in a fluid
$W_{\text{app}} = W - F_b$
Float or sink — the density comparison
Fraction submerged of a floating body
$\dfrac{V_{\text{submerged}}}{V_{\text{total}}} = \dfrac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$

Watch out for (6)

Relative density has no unit
→ Density and relative density
Equal volumes vs equal masses give DIFFERENT averages
→ Combining densities — equal volumes vs equal masses
Upthrust = weight of displaced FLUID, not of the body
→ Archimedes' principle and the buoyant force
Mass is unchanged; only apparent weight drops
→ Apparent weight loss on submersion
It is AVERAGE density that decides flotation
→ Float or sink — the density comparison
Stability is about M above G, not G below B
→ Stability of a floating body — the metacentre

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Fluid Mechanics and Properties of MatterHARD

Two identical containers X and Y are connected at the bottom by a thin tube with a valve. X has liquid at height h, Y is empty. When valve is opened, both containers have equal liquid. If initial PE of liquid is

P_1

and final PE is

P_2

, then :

[Q54 · Apr · 2023]

Example 2Fluid Mechanics and Properties of MatterMODERATE

An object is made of two equal parts by volume; one part has density

\rho_0

and the other part has density

2\rho_0

. What is the average density of the object?

[Q63 · Sep · 2022]

Example 3Fluid Mechanics and Properties of MatterEASY

Buoyancy is a/an

[Q114 · Sep · 2021]

Example 4Fluid Mechanics and Properties of MatterHARD

Shown in the figure are two hollow cubes C1 and C2 of negligible mass partially filled with liquids of densities

\rho

1 and

\rho

2 respectively, floating in water (density

\rho

W). The relationship between

\rho

\rho

2 and

\rho

W is

[Q64 · Sep · 2024]

Example 5Fluid Mechanics and Properties of MatterEASY

Which one of the following regarding density of water at atmospheric pressure is correct ?

[Q52 · Apr · 2021]

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