Upthrust and Floatation - Vardaan Learning Institute

Created by Team Vardaan with Powered by

ICSE Class 9 Physics • Chapter 5 • Comprehensive Chapter Notes

(A) Upthrust and Archimedes' Principle

5.1 Buoyancy and Upthrust

Fig. - Buoyancy and Upthrust

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A split panel. Left Panel: A human hand pushing an empty sealed tin can downwards into a transparent tub of blue water, with upward translucent ripples indicating resistance. Right Panel: A solid block perfectly submerged in blue water, showing a distinct red downward arrow labeled 'W' (Weight) starting from its center, and a distinct blue upward arrow labeled 'F_B' (Upthrust) acting upwards from the same center. Minimalist and strictly academic. IMPORTANT: You must include professional mathematical text labels. On the block, label the red downward arrow as "Weight W = mg" and the blue upward arrow as "Upthrust F_B". Do not clutter.

Concept

When a body is partially or wholly immersed in a liquid, an upward force acts on it. This upward force is known as upthrust or buoyant force ($F_B$).

Buoyancy: The property of a liquid to exert an upward force on a body immersed in it.

Unit: Upthrust, being a force, is measured in newton (N) or kgf.

Experimental Demonstrations:

Pushing an empty can into water: An empty sealed can floats on water. Pushing it down requires force. The further it is pushed, the greater the upward force felt. If released, it bounces back to the surface.
Pushing a cork into water: A cork floats with roughly 2/5th of its volume inside. Pushing it inside requires downward force, and it rises back up when released due to buoyant force.

Fact

Condition for a body to float or sink: When a body is immersed in a fluid, two forces act: (i) Weight ($W$) acting vertically downwards. (ii) Upthrust ($F_B$) acting vertically upwards.

If $F'_B > W$ or $F'_B = W$, the body will float. ($F'_B$ is max upthrust when fully immersed).
If $F'_B < W$, the body will sink.

Note: Gases also exert upthrust. A hydrogen balloon rises because the upthrust of the surrounding air is greater than the balloon's weight.

5.2 Characteristic Properties of Upthrust

Fig. - Center of Buoyancy

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A block floating in blue liquid with exactly half of its volume submerged. A red dot labeled 'G' (Center of Gravity) is perfectly in the geometric center of the entire block. A blue dot labeled 'B' (Center of Buoyancy) is positioned lower, exactly at the geometric center of the *submerged* portion of the block. A red arrow points down from G, and a blue arrow points up from B. Minimalist, precise physics visualization. IMPORTANT: You must include professional mathematical text labels. Label the red dot "G (Center of Gravity)" and its downward arrow "W = mg". Label the blue dot "B (Center of Buoyancy)" and its upward arrow "F_B". Do not clutter.

Larger the volume of body submerged, greater is the upthrust: A bunch of feathers and a pebble of the same mass fall differently in air. The pebble falls faster because its volume is smaller, so it experiences less upthrust from air than the larger bunch of feathers.
For the same volume inside the fluid, more the density of fluid, greater is the upthrust: A cork requires more downward force to be completely immersed in glycerine (denser) than in water (less dense).
The upthrust acts upwards at the centre of buoyancy: The centre of buoyancy is the centre of gravity of the displaced fluid (i.e., the submerged part of the body). For a fully immersed uniform body, it coincides with the centre of gravity of the body.

5.3 Reason for Upthrust

Fig. - Reason for Upthrust

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A transparent cylindrical body is perfectly submerged vertically inside a blue liquid. Downward short red arrows point to the top circular face (labeled P1). Upward long red arrows point to the bottom circular face (labeled P2), indicating a pressure difference due to depth. Horizontal arrows on the curved sides are equal and opposing each other to show lateral cancellation. Clean, academic layout. IMPORTANT: You must include professional mathematical text labels. Label the top face "Pressure P₁ = h₁ρg", the bottom face "Pressure P₂ = h₂ρg". Add an equation "P₂ > P₁" and label the net upward force as "Upthrust F_B = (P₂ - P₁)A".

A liquid exerts pressure at all points. Pressure increases with depth. When a body is immersed in a liquid, the pressure $P_2$ exerted upwards on its lower face (at greater depth) is strictly greater than the pressure $P_1$ exerted downwards on its upper face (at lesser depth). The difference in pressure $(P_2 - P_1)$ creates a net upward force, which is the upthrust.

Note: Thrust on the side walls neutralises as they are equal and opposite. A thin lamina experiences negligible upthrust as upper and lower depths are practically the same.

5.4 Upthrust is Equal to the Weight of Displaced Liquid (Mathematical Proof)

DERIVATION

Consider a cylindrical body PQRS of cross-sectional area $A$ immersed in a liquid of density $\rho$. Let top surface PQ be at depth $h_1$ and bottom RS at depth $h_2$.

Downward thrust on top surface: $F_1 = P_1 \times A = h_1 \rho g A$

Upward thrust on bottom surface: $F_2 = P_2 \times A = h_2 \rho g A$

Resultant upward thrust (Buoyant Force):

$$ F_B = F_2 - F_1 = (h_2 - h_1)\rho g A $$

Since $A(h_2 - h_1) = V$ (volume of submerged part of body):

$$ F_B = V \rho g $$

Since $V \rho g = \text{mass of displaced liquid} \times g = \text{Weight of displaced liquid}$.

Upthrust = Weight of the liquid displaced by the submerged part of the body.

Effect of Upthrust: A body appears to weigh less in a liquid than its actual weight. For example, lifting a bucket of water from a well feels easier as long as the bucket is inside the water.

5.5 Archimedes' Principle

Statement ICSE "When a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of the liquid displaced by it."

5.6 Experimental Verification of Archimedes' Principle

Fig. - Eureka Can Verification

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. An Archimedes principle verification setup. A metallic cylindrical weight is suspended from a vertical mechanical spring balance. The weight is completely submerged in a water-filled Eureka can (displacement can). The displaced water is flowing out from the can't side spout and collecting neatly into a small glass measuring cylinder resting next to it. Clean, academic 3D vector style. IMPORTANT: You must include professional mathematical text labels. Label the spring balance reading as "Apparent Weight". Label the displaced water in the cylinder as "Weight of Displaced Liquid". Add an equation label "Upthrust = Weight of Displaced Liquid".

Experiment (Eureka Can): Suspend a solid from a spring balance and note its weight in air (e.g., 300 gf).
Fill a Eureka can up to its spout and place a measuring cylinder under it.
Immerse the solid completely in the can. The spring balance reads less (e.g., 200 gf), meaning loss in weight = 100 gf.
The volume of water displaced into the cylinder will be $100 \text{ cm}^3$. Since density of water is $1 \text{ g cm}^{-3}$, the weight of displaced water is 100 gf.
Conclusion: Loss in weight (Upthrust) = Weight of water displaced.

5.7 Floatation based on Density

Fig. - Density Float/Sink Demo

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A large transparent glass beaker filled with blue water. An iron nail is shown resting entirely on the bottom of the beaker (sinking). A brown wooden cork is shown floating happily at the very top surface of the water, partially submerged. Minimalist academic demonstration. IMPORTANT: You must include professional mathematical text labels. Next to the sunken nail, add label "Density ρ_solid > ρ_liquid (Sinks)". Next to the floating cork, add label "Density ρ_solid < ρ_liquid (Floats)". Add downward "W" and upward "F_B" vectors on both objects.

Let a body of volume $V$ and density $\rho$ be fully immersed in a liquid of density $\rho_L$. Weight $W = V \rho g$, Max upthrust $F'_B = V \rho_L g$.

Conditions

If $\rho > \rho_L$, $W > F'_B$. The body will sink. (e.g., Iron nail in water).
If $\rho = \rho_L$, $W = F'_B$. The body will float fully submerged. Net force is zero.
If $\rho < \rho_L$, $W < F'_B$ (if forced to submerge). The body will float with a part of it submerged such that upthrust equals weight. (e.g., Cork in water).

✍ IN-TEXT PRACTICE (NUMERICALS)

Q1. A body weighs 200 gf in air and 190 gf when completely immersed in water. Calculate (i) loss in weight, (ii) upthrust.

(i) Loss in weight: $200 \text{ gf} - 190 \text{ gf} = \mathbf{10 \text{ gf}}$

(ii) Upthrust: Upthrust = Loss in weight = $\mathbf{10 \text{ gf}}$

Q2. A piece of iron of density $7.8 \times 10^3 \text{ kg m}^{-3}$ and volume $100 \text{ cm}^3$ is completely immersed in water ($\rho = 1000 \text{ kg m}^{-3}$). Calculate: (i) weight of iron piece in air, (ii) upthrust, (iii) apparent weight in water. ($g = 10 \text{ m s}^{-2}$)

Given: $V = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3$

(i) Weight in air: $W = V \rho_{\text{iron}} g = 10^{-4} \times (7.8 \times 10^3) \times 10 = \mathbf{7.8 \text{ N}}$

(ii) Upthrust: $F_B = V \rho_{\text{water}} g = 10^{-4} \times 1000 \times 10 = \mathbf{1 \text{ N}}$

(iii) Apparent Weight: $W_{\text{app}} = W - F_B = 7.8 - 1 = \mathbf{6.8 \text{ N}}$

(B) Relative Density and Its Measurement by Archimedes' Principle

5.8 Density

The density of a substance is its mass per unit volume.

FORMULA $$ \text{Density } (\rho) = \frac{\text{Mass } (M)}{\text{Volume } (V)} $$

Units: S.I. unit is kg m$^{-3}$. C.G.S. unit is g cm$^{-3}$.

Conversion: $1 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$.

Effect of Temperature: Density decreases with heating (expansion) and increases with cooling. Exception: Water. The density of water is maximum at $4^{\circ}\text{C}$, equal to $1 \text{ g cm}^{-3}$ or $1000 \text{ kg m}^{-3}$.

5.9 Relative Density (R.D.)

Definition

The relative density (R.D.) of a substance is the ratio of the density of that substance to the density of water at $4^{\circ}\text{C}$.

$$ R.D. = \frac{\text{Density of substance } (\rho_s)}{\text{Density of water at } 4^{\circ}\text{C } (\rho_w)} $$

Alternatively: $R.D. = \frac{\text{Mass of substance}}{\text{Mass of an equal volume of water at } 4^{\circ}\text{C}}$

Unit: R.D. is a pure ratio and has no unit.

Relationship: Density in $\text{g cm}^{-3}$ is numerically equal to R.D. Density in $\text{kg m}^{-3} = (R.D.) \times 1000$.

5.10 & 5.11 Determination of R.D. by Archimedes' Principle

Fig. - Measuring Relative Density

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A traditional dual-pan physical laboratory balance. The left pan holds standard metallic weights. From the right hook, a solid dark weight is suspended by a string. This solid is perfectly submerged in a glass beaker of water. Crucially, the beaker is resting independently on a small wooden bridge that straddles *over* the balance pan without touching the pan itself. Professional laboratory setup, clean, strictly academic. IMPORTANT: You must include professional mathematical text labels. Label the solid suspended in water as "Weight in Water (W₂)". Label the weights on the other pan as "Apparent Weight". Add an equation label "Loss of Weight = W₁ - W₂" pointing to the submerged solid.

MEASUREMENT

R.D. of a Solid (denser than water, insoluble):

$$ R.D. = \frac{\text{Weight of solid in air}}{\text{Loss of weight in water}} = \frac{W_1}{W_1 - W_2} $$

(If soluble in water, weigh in a liquid where it is insoluble, and multiply the ratio by R.D. of that liquid).

R.D. of a Liquid:

$$ R.D. = \frac{\text{Loss of weight of a solid in liquid}}{\text{Loss of weight of same solid in water}} = \frac{W_1 - W_2}{W_1 - W_3} $$

Where $W_1$ is weight of solid in air, $W_2$ in liquid, and $W_3$ in water.

✍ IN-TEXT PRACTICE

Q. A solid weighs 50 gf in air and 44 gf when completely immersed in water. Calculate (i) upthrust, (ii) volume of solid, (iii) R.D. of solid.

(i) Upthrust: $W_1 - W_2 = 50 - 44 = \mathbf{6 \text{ gf}}$

(ii) Volume: Weight of water displaced = 6 gf. Density of water = $1 \text{ g cm}^{-3}$, so volume displaced = $6 \text{ cm}^3$. Therefore, volume of solid = $\mathbf{6 \text{ cm}^3}$.

(iii) R.D.: $\frac{W_1}{W_1 - W_2} = \frac{50}{6} = \mathbf{8.33}$

(C) Floatation

5.12 Principle of Floatation

Fig. - Cases of Floatation

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. Three identical glass beakers side-by-side, each filled with blue liquid. Beaker 1: A solid block sunken entirely to the bottom floor. Beaker 2: A solid block floating completely submerged, hovering just below the liquid surface. Beaker 3: A solid block floating partially above the liquid surface. Clean academic physics diagram. IMPORTANT: You must include professional mathematical text labels. Beaker 1: label "W > F_B (Sinks)". Beaker 2: label "W = F_B (Submerged)". Beaker 3: label "W = F_B (Partially Submerged)". Include vector arrows for W and F_B on each block.

Core Principle ICSE "According to the principle of floatation, the weight of a floating body is equal to the weight of the liquid displaced by its submerged part." $$ W = F_B $$

Therefore, the apparent weight of a floating body is zero.

5.13 Relation Between Volume of Submerged Part and Densities

Let $V$ = total volume, $v$ = submerged volume, $\rho_s$ = density of solid, $\rho_L$ = density of liquid.

Weight of body $W = V \rho_s g$. Upthrust $F_B = v \rho_L g$. For floatation, $W = F_B$:

RELATION $$ V \rho_s g = v \rho_L g \implies \frac{v}{V} = \frac{\rho_s}{\rho_L} $$

Volume of immersed part / Total volume = Density of body / Density of liquid.

Example: An iceberg ($\rho = 0.917 \text{ g cm}^{-3}$) in water ($\rho = 1 \text{ g cm}^{-3}$) will have $\frac{v}{V} = \frac{0.917}{1} = 91.7\%$ of its volume submerged.

5.14 Applications of the Principle of Floatation

Fig. - Floatation Applications

Diagram Description: A highly professional, landscape academic diagram on a pure white (#ffffff) background. A triple-panel view of floatation applications. Panel 1: A massive metallic cargo ship floating stably on water. Panel 2: A submarine submerged with its periscope above water, showing internal red-shaded ballast tanks. Panel 3: A massive iceberg floating with roughly 90% of its bulk hidden deep underwater. Clean, professional 3D vector style. IMPORTANT: You must include professional text labels. Panel 1: Label the ship with "Average Density < Water". Panel 2: Label submarine ballast tanks with "Variable Density". Panel 3: Label the iceberg with "ρ_ice < ρ_water" and "90% Submerged". Include vector arrows for "W" and "F_B" balancing each other.

Applications

Floatation of iron ship: An iron nail sinks, but a ship is hollow. Its average density is less than water. A loaded ship submerses more than an unloaded one. A ship sailing from sea water (denser) to river water (less dense) will sink further to displace more volume. Plimsoll line indicates safe loading limit.
Floatation of human body: Average density with lungs filled is $\sim 1 \text{ g cm}^{-3}$. It is easier to swim in sea water (density $1.026 \text{ g cm}^{-3}$) than fresh water, as less of the body needs to be submerged to balance weight.
Floatation of submarines: Possesses ballast tanks. Filling them with water increases average density, making it dive. Pushing water out with compressed air decreases density, making it rise.
Floatation of iceberg: Ice density ($0.917 \text{ g cm}^{-3}$) is less than water. It floats with $\sim 90\%$ inside, making it dangerous for ships. Note: When floating ice melts, the water level remains unchanged because ice contracts on melting.
Floatation of fish: Uses a "swim bladder" to change its volume and average density to rise or sink.
Rising of balloons: A balloon filled with hydrogen (less dense than air) rises because upthrust of air > weight of balloon. It stops rising when air density drops at high altitudes such that upthrust equals weight.