JEE/NEET Physics · Mechanical Properties of Fluids series · Part 3 of 7 · All parts →
- Archimedes: upthrust = weight of displaced fluid, F_B = ρ_fluid × V_displaced × g
- Float if density < fluid's; sink if greater; hang submerged if equal
- Floating body displaces exactly its own weight of fluid
- Apparent weight in fluid = true weight − upthrust
- Ice floats with ~90% submerged (ρ_ice/ρ_water ≈ 0.9); ships ride higher in salt water
A 200,000-tonne ship floats; a 10-gram coin sinks. Weight has nothing to do with it — density decides everything. Archimedes figured this out in a bathtub and ran through the streets shouting. Part 3 of the Mechanical Properties of Fluids series.
- The upthrust idea
- Archimedes’ principle
- Float, sink, or hover
- Apparent weight
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Recap
The Upthrust Idea
Water pushes harder on the bottom of a submerged object than on its top (pressure grows with depth). The net of these pushes is a single upward force — buoyancy, the fluid’s attempt to reclaim its space.
Archimedes’ Principle
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| F_B | buoyant force (upthrust) | N, always upward |
| ρ_fluid | density of the FLUID (not the object!) | kg/m³ |
| V_submerged | volume of object under the surface | m³ |
Float, Sink, or Hover
| Density comparison | Result | Example |
|---|---|---|
| ρ_obj < ρ_fluid | floats, partially submerged | wood, ice, ships |
| ρ_obj = ρ_fluid | hovers at any depth | fish with adjusted bladder |
| ρ_obj > ρ_fluid | sinks, but lighter while sinking | stone, coin, iron |
Apparent Weight
Submerged, an object’s scale reading drops by the upthrust: W_app = mg − ρ_f V g. This loss is exactly the weight of displaced water — and it’s how density is measured by the immersion method.
Solved Examples
V = m/ρ = 1/2500 = 4×10⁻⁴ m³; F_B = 1000 × 4×10⁻⁴ × 10 = 4 N.
W_app = 10 − 4 = 6 N.
✔
Answer: 6 N
Weight = upthrust: 900×V_total×g = 1000×V_sub×g.
V_sub/V_total = 900/1000 = 90% submerged — the tip of the iceberg is literally 10%. ✔
Answer: 90% below
Weight = sum of upthrists: ρ_c×V×g = 1000×(V/2)×g + 800×(V/2)×g.
ρ_c = (1000+800)/2 = 900 kg/m³.
✔
Answer: 900 kg/m³
- Using the object’s density in F_B. Upthrust involves the FLUID’s density — the object’s density only sets its weight.
- Submerged volume confusion. F_B uses only the volume UNDER the surface; a floating ship displaces its weight, not its volume, of water.
- ‘Heavy things sink.’ Density decides, not mass: a 200,000-tonne ship is a hollow object with average density less than water’s.
- Sign errors in apparent weight. W_app = mg − F_B, always a subtraction when submerged.
This Physics in Your Daily Life
- Why you float in the Dead Sea — its ~1240 kg/m³ brine beats your body’s ~1000: reading in a bathtub, buoyancy is literally denser water.
- Submarines dive and rise by flooding or blowing ballast tanks: adjusting average density at will — engineered hover states.
- Hot-air balloons and helium parties — buoyancy in air: the displaced air outweighs the warm gas inside.
- Hydrometers checking car batteries and milk purity — float depth directly reads density: Archimedes as an instrument.
- Life jackets and pool noodles — foam’s low density drags YOUR average density below water’s: safety as density engineering.
Replace the submerged object with water of exactly the same shape: that water would hang in perfect balance (it’s the same as the water around it). What forces balanced it? Its own weight — supplied by the pressure field. Now put the object back: the pressure field hasn’t changed, so it supplies the same push: the displaced water’s weight.
A 1-litre bottle submerged: upthrust = 1000×0.001×10 = 10 N (about 1 kg of water pushed aside). Fill it with sand (3 kg): net downward 20 N — sinks. Empty (0.2 kg): net up 8 N — bobs up until only 0.2 litres sits under.
Draw the submerged block with pressure arrows: short ones on top, long ones below, equal sideways. Add them vectorially: sides cancel, and the leftover is a single upward arrow — the pressure triangle’s vote. Its size works out to ρ_fVg exactly.
Practice set (answers hidden — try first)
(NEET-level) 2 kg, ρ=4000, in water (g=10): F_B =
(JEE Main-level) Object’s apparent weight in water is 3/4 of true: density =
(NEET-level) Ice (900) in water: submerged fraction =
(Concept) A floating ship displaces water equal to its:
(JEE Main-level) Block floats 40% in a liquid of 1250: block density =
- F_B = ρ_fluid V g — fluid’s density, submerged volume
- density comparison decides float/sink
- floaters displace their own WEIGHT of fluid
- W_app = mg − F_B
- ice: ~90% under, 10% showing
- 🔁 upthrust = displaced weight
- 🔁 fluid density (not object’s) in F_B
- 🔁 float/sink/hover table
- 🧠 Chant: ‘push aside water, water pushes back its weight’.
- 🧠 Decider: ‘average density vs the fluid’s’.
- 🏠 Daily: Dead Sea floating = denser water.
- 🏠 Daily: life jackets engineer your density.
Quick revision
- Archimedes: upthrust = weight of displaced fluid, F_B = ρ_fluid × V_displaced × g
- Float if density < fluid's; sink if greater; hang submerged if equal
- Floating body displaces exactly its own weight of fluid
- Apparent weight in fluid = true weight − upthrust
- Ice floats with ~90% submerged (ρ_ice/ρ_water ≈ 0.9); ships ride higher in salt water
- Archimedes’ principle
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