Buoyancy and Archimedes' Principle

Buoyancy is one of those topics where students think they understand it after one reading and then lose marks on every second question. The reason is simple — Archimedes’ principle is short, but the situations where you apply it are subtle. We will work through the concept the way it should have been taught the first time.

This chapter shows up in CBSE Class 9, Class 11 (Mechanical Properties of Fluids), and is high-weightage in NEET. JEE Main occasionally throws a buoyancy + spring or buoyancy + accelerating frame problem that catches everyone off guard.

What Buoyancy Actually Is

When an object is submerged (fully or partially) in a fluid, the fluid pushes on every surface of the object. The pressure at the bottom is higher than at the top because pressure increases with depth. This pressure difference produces a net upward force — that’s the buoyant force.

It has nothing to do with the object’s material. A steel needle and a steel ship displace different volumes of water, so they experience different buoyant forces.

The buoyant force is a property of the displaced fluid, not the object. Two completely different objects displacing the same volume of the same fluid feel the same buoyant force.

Archimedes’ Principle

The buoyant force on a body submerged in a fluid equals the weight of the fluid displaced by the body.

F_B = \rho_{fluid} \cdot V_{displaced} \cdot g

Here $\rho_{fluid}$ is the density of the fluid (not the object), and $V_{displaced}$ is the volume of fluid pushed out of the way by the submerged part of the object.

For a fully submerged object, $V_{displaced}$ equals the object’s full volume. For a floating object, $V_{displaced}$ is the volume of the submerged portion only.

Three Cases of Submersion

Case 1: Object Sinks

Happens when $\rho_{object} > \rho_{fluid}$ . The object is fully submerged, and the buoyant force is less than the weight. Apparent weight in fluid:

W_{apparent} = W - F_B = (\rho_{obj} - \rho_{fluid}) V g

Case 2: Object Floats

Happens when $\rho_{object} < \rho_{fluid}$ . The object floats with part of it above the fluid. Equilibrium:

\rho_{obj} V g = \rho_{fluid} V_{sub} g

So the fraction submerged equals $\rho_{obj}/\rho_{fluid}$ . For ice in water: $0.92/1 = 92\%$ submerged. That is why icebergs are mostly underwater.

Case 3: Object in Equilibrium Inside the Fluid

Happens when $\rho_{object} = \rho_{fluid}$ . The object can stay at any depth — neutral buoyancy. Submarines achieve this by adjusting ballast.

Solved Examples

Example 1 (Easy) — CBSE Class 9 Style

A wooden block of mass $200$ g floats in water with $80\%$ of its volume submerged. Find the volume of the block.

Equilibrium: weight = buoyant force on submerged part.

0.2 \, \text{kg} \times 10 = 1000 \times 0.8 V \times 10

$V = 0.2/(8000) = 2.5 \times 10^{-4}$ m $^3$ $= 250$ cm $^3$ .

Example 2 (Medium) — JEE Main

A cubical block of side $10$ cm and density $800$ kg/m $^3$ is held submerged in water by a string. Find the tension in the string.

Volume $V = 10^{-3}$ m $^3$ . Weight = $800 \times 10^{-3} \times 10 = 8$ N. Buoyancy = $1000 \times 10^{-3} \times 10 = 10$ N. Tension = $F_B - W = 2$ N (string pulls down to keep block submerged).

Example 3 (Hard) — JEE Advanced

A beaker of water sits on a weighing scale. A solid metal ball of mass $500$ g and density $5000$ kg/m $^3$ is suspended by a string and lowered into the water until fully submerged (without touching the bottom). What is the new reading of the scale?

Volume of ball $= 500/5000 = 0.1$ litre $= 10^{-4}$ m $^3$ . Buoyancy on ball = $1000 \times 10^{-4} \times 10 = 1$ N. By Newton’s third law, the ball pushes the water down by $1$ N. The scale reads original weight + $1$ N $= W_{original} + 0.1$ kg equivalent.

If the original beaker+water weighed $2$ kg, scale now reads $2.1$ kg.

JEE Advanced 2022 had this exact “suspended ball + scale reading” question. Most students forgot Newton’s third law and answered “no change”. The correct answer needs the buoyant reaction force on the water.

Buoyancy in Accelerating Frames

This is the JEE Advanced trap. In a frame accelerating with acceleration $a$ (upward, say), effective gravity becomes $g_{eff} = g + a$ .

F_B = \rho V (g + a)

If the lift is in free fall ( $a = -g$ ), buoyancy is zero — objects neither sink nor float. A balloon released inside a free-falling lift just hangs there.

In a horizontally accelerating tank, the water surface tilts, and the buoyancy direction tilts too. The effective $g$ is $\sqrt{g^2 + a^2}$ at an angle.

Pressure and Buoyancy

Why does a denser fluid produce more buoyancy? Because pressure at depth $h$ is $\rho g h$ . More density → more pressure difference top-to-bottom → more upward push.

For a submerged cylinder of cross-section $A$ and height $h$ :

F_B = (P_{bottom} - P_{top}) A = \rho g h \cdot A = \rho g V

Same formula, derived from pressure. Useful when the shape is irregular — replace $V$ with the actual submerged volume.

Common Mistakes

Mistake 1: Using object’s density instead of fluid’s in the buoyancy formula. Always $\rho_{fluid}$ .

Mistake 2: For a floating object, plugging in the full volume instead of only the submerged volume.

Mistake 3: Forgetting that in non-inertial frames, $g$ becomes $g_{eff}$ .

Mistake 4: Treating buoyancy as the net force — it is one of two forces (the other being weight) acting on the object.

Mistake 5: When a block partially floats and is then pressed deeper, students forget to update $V_{sub}$ and the new equilibrium condition.

Practice Questions

Q1. A 5 kg block of density 2500 kg/m³ is fully submerged in water. Find the buoyant force on it.

Volume $= 5/2500 = 0.002$ m $^3$ . $F_B = 1000 \times 0.002 \times 10 = 20$ N.

Q2. Ice floats in water with 92% submerged. What fraction of an iceberg is above the water surface in seawater (density 1025 kg/m³)?

Submerged fraction $= 920/1025 \approx 0.898$ . Above water $\approx 10.2\%$ .

Q3. A boy floats with face above water. The water in the swimming pool is replaced by mercury. What happens?

Mercury density $13.6$ times water. The boy floats much higher — only a few percent of his body submerged. Possibly only feet touch.

Q4. A balloon filled with helium is released in air. Why does it rise, and what determines its terminal height?

Buoyant force from displaced air exceeds the balloon + helium weight. As it rises, air becomes less dense; equilibrium is reached when air density equals the balloon’s effective density.

Q5. A wooden cube floats with 60% submerged in water. What density of liquid would make it float with 80% submerged?

Wood density $= 0.6 \times 1000 = 600$ kg/m $^3$ . New liquid density: $\rho = 600/0.8 = 750$ kg/m $^3$ .

Q6. Find the buoyancy force on a sphere of radius 5 cm fully submerged in oil of density 800 kg/m³.

Volume $= \frac{4}{3}\pi (0.05)^3 \approx 5.24 \times 10^{-4}$ m $^3$ . $F_B = 800 \times 5.24 \times 10^{-4} \times 10 \approx 4.19$ N.

Q7. A boat carries a stone. The stone is dropped into the lake. Does the water level rise, fall, or stay the same?

Falls. While in the boat, stone displaces water equal to its weight. Once submerged, it displaces only its volume — which is less (stone denser than water).

Q8. A cube of side 10 cm and density 700 kg/m³ floats in water. How much weight must be placed on top to fully submerge it?

Cube volume $10^{-3}$ m $^3$ . Cube weight $= 7$ N. Max buoyancy when fully submerged $= 10$ N. Extra weight needed $= 3$ N $= 0.3$ kg.

FAQs

Why do steel ships float but a steel needle sinks? The ship is hollow and displaces a lot of water; its average density (steel + air inside) is less than water. The needle’s density is the steel’s density, much greater than water.

Does Archimedes’ principle work in air? Yes. Hot-air balloons and helium balloons demonstrate buoyancy in air. Air’s density is small (~1.2 kg/m³), so the effect is noticeable only for low-density gases or large volumes.

What happens to buoyancy if the fluid is rotating? In a rotating frame, centrifugal acceleration adds to gravity along the rotation axis; the surface curves into a parabola, and effective $g$ varies with radius.

Can buoyancy be horizontal? Yes — in horizontally accelerating fluids, the “effective gravity” tilts, and so does the buoyant force.

Why does a hot-air balloon stop rising at some altitude? As it rises, air density decreases; eventually the air’s density equals the balloon’s average density and buoyancy equals weight.

Does the shape of the object matter for buoyancy? Only the submerged volume matters, not the shape. A flat plate and a sphere with the same submerged volume feel the same buoyant force.