Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation
More than two thousand years ago, in the Greek city of Syracuse in Sicily, a king named Hiero II ordered a crown of pure gold. When the crown came back from the goldsmith, it weighed exactly as much as the gold the king had handed over. Yet a suspicion remained. Had the goldsmith kept some of the gold and mixed in cheaper silver instead? The crown could not be melted down or cut open. The king turned to Archimedes.
The difficulty was not weighing the crown. That was easy. The difficulty was measuring the volume of an object with such an awkward shape, because gold and silver of the same weight take up different amounts of space. According to the account written down by the Roman writer Vitruvius about two centuries later, the answer came to Archimedes as he lowered himself into a full bath and watched the water spill over the edge. The water pushed aside by his body had exactly the same volume as the part of him that was under the surface. A shape too complicated to measure with a ruler could be measured with water.
That single observation grew into a rule about the upward push a liquid gives to anything placed in it. You use that rule every time you lift a bucket from a well, watch a ship ride at anchor, or float on your back in a swimming pool.
Diagram
Fig. 5.0-1 — Measuring The Volume Of An Irregular Object By Displacement Two side-by-side sketches. Left: an ornate crown resting on a pan of a simple beam balance, with an equal-sized lump of gold on the other pan, beam level; label ‘equal weight’. Right: a tall jar filled to the brim standing inside a wide catch-tray, with the crown being lowered into it on a thread and water spilling over the rim into the tray; label the jar ‘water filled to the brim’, label the spilled water ‘water pushed aside has the same volume as the crown’, and label the thread ‘crown lowered gently’.
5.1 Buoyancy And Upthrust
Lower your hand slowly into a bucket of water and you will feel the water pressing back. Push a closed, empty bottle down into the same bucket and the push-back is unmistakable: the water shoves the bottle upwards, and the deeper you take it, the harder it shoves. Every liquid does this, and so does every gas.
Anything that can flow — a liquid or a gas — is called a fluid. When a body is partly or fully placed inside a fluid, the fluid pushes it upwards. This upward force is called the upthrust, or the buoyant force, and it is written with the symbol \(F\_B\).
The property of a fluid by which it exerts this upward force on a body placed in it is called buoyancy. Upthrust is the force itself; buoyancy is the property of the fluid that produces it.
5.1.1 Showing Upthrust By Experiment
Two simple experiments make the upthrust easy to feel. The first is drawn in Fig. 5.1-1.
Experiment 1 — pushing an empty can into water. Take an empty can and close its mouth with an airtight stopper so that no water can get in. Place it in a tub of water. It floats with a large part of it above the water surface and only a small part below.
Now push the can down into the water. You feel an upward force opposing your push, and you find it difficult to press the can further in. The more of the can you force under the surface, the more force you need. This continues until the can is completely immersed. Even then a steady force is still needed to hold it in place, and this is the largest upthrust the water can give that can. Let the can go, and it springs back to the surface and floats again.
Diagram
Fig. 5.1-1 — Pushing A Closed Empty Can Into Water A tub of water with an empty can, mouth closed by a stopper, being pushed below the surface by a hand pressing down on it. Draw an arrow pointing vertically down from the hand labelled ‘push’, an arrow pointing vertically up from the centre of the can labelled ‘upthrust F_B’, and an arrow pointing vertically down labelled ‘weight W’. Show the water surface as a horizontal line and mark the part of the can below the surface with light shading labelled ‘submerged part’.
Experiment 2 — pushing a cork into water. A piece of cork placed on the surface of water in a tub floats with only a small part of its volume below the surface. Push the cork under and release it, and it rises and floats again. While you hold it under, your fingers feel a steady upward push. The cork behaves exactly like the closed can.
Explaining What You Felt
Two forces act on the can or the cork in the tub. One is its weight \(W\), the force of gravity pulling it downwards. The other is the upthrust \(F\_B\) of the water, pushing it upwards. The body settles and floats in the position where these two forces are equal in size, that is, where
\[W = F\_B\]
As the can is pushed further and further in, more of its volume is inside the water, and the upthrust \(F\_B\) grows. It reaches its largest value, written \(F'\_B\), when the can is completely immersed. Since this maximum upthrust \(F'\_B\) is now greater than the weight \(W\) of the can, releasing the can leaves an unbalanced upward force, and the can rises. To keep it fully immersed, you must supply an extra downward force equal to \(F'\_B - W\) to balance that unbalanced upward push.
Liquids are not the only fluids that behave this way. Gases do too, so a body placed in a gas also experiences an upthrust. All of us are being pushed upwards at this moment by the air around us, but we never notice it, because that upthrust is extremely small compared with our own weight. A balloon filled with hydrogen, or with any gas less dense than air, does rise, because the upthrust on it from the surrounding air is more than the weight of the balloon together with the gas inside it.
5.1.2 Apparent Weight
A body inside a fluid has two forces on it: its real weight acting downwards, and the upthrust of the fluid acting upwards. The upward push cancels part of the downward pull, so the body seems lighter than it really is. The weight a body seems to have while it is in a fluid is called its apparent weight.
\[\\text{apparent weight} = \\text{real weight} - \\text{upthrust}\]
where the real weight is the weight of the body in air, the upthrust is the upward force of the fluid, and all three are forces measured in newton (N). Notice that the body has not lost any matter. Its mass is unchanged; only the net downward force on it has been reduced.
A bucket being drawn from a well shows this clearly. While the bucket is still below the water, it is easy to pull. The moment it rises clear of the water surface, it suddenly feels much heavier, because the upthrust of the water is no longer helping you. For the same reason, a fish lifted out of water feels heavier than it did while it was still in the water.
5.1.3 Condition For A Body To Float Or Sink In A Fluid
We have seen that the upthrust grows as more of the body goes under, and becomes largest, \(F'\_B\), when the body is completely immersed. Fig. 5.1-2 shows a body held completely immersed in a fluid, with the weight \(W\) acting downwards and the maximum upthrust \(F'\_B\) acting upwards.
Diagram
Fig. 5.1-2 — Forces On A Body Held Completely Immersed In A Liquid A vessel of liquid shown by horizontal dashes, with a rectangular block drawn fully below the liquid surface. From the centre of the block draw one arrow vertically upwards labelled ‘F′_B (maximum upthrust)’ and one arrow vertically downwards labelled ‘W (weight)’. Mark the free surface of the liquid at the top.
Whether the body finally floats or sinks depends on which of these two forces is larger, and that depends on the density of the fluid. Three outcomes are possible.
- If \(F'\_B > W\), the body will not stay under. It rises and floats partly immersed, with just enough of it inside the liquid for the upthrust \(F\_B\) on the submerged part to become equal to the weight \(W\) of the body.
- If \(F'\_B = W\), the body floats with the whole of it just immersed inside the liquid, as in Fig. 5.1-2.
- If \(F'\_B < W\), the body sinks, because even the largest upthrust the fluid can offer cannot balance the weight. The unbalanced downward force on it is \(W - F'\_B\).
In both of the first two cases the body floats, and the net downward force on it — that is, its apparent weight — is zero. This is worth holding on to: a floating body has zero apparent weight.
In the third case the body accelerates downwards through the liquid. If \(m\) is the mass of the body, then by Newton’s second law the unbalanced force produces an acceleration \(a\) given by
\[ma = W - F'\_B \\qquad \\text{or} \\qquad a = \\frac{W - F'\_B}{m}\]
where \(W\) and \(F'\_B\) are in newton (N), \(m\) is in kilogram (kg) and \(a\) is in \(\\text{m s}^{-2}\). This expression ignores the viscous force of the liquid, that is, the resistance the liquid offers to a body moving through it.
5.1.4 Unit Of Upthrust
Upthrust is a force, so it is measured in the same units as any other force. Its SI unit is the newton (N). It is also often quoted in the gravitational unit kilogram-force (kgf), where one kgf is simply the weight of a body of mass one kilogram. In the same way, one gram-force (gf) is the weight of a body of mass one gram.
5.2 Characteristic Properties Of Upthrust
The upthrust of a fluid is not a vague push. It follows three definite rules, and these three between them decide everything else in this chapter.
- The larger the volume of a body submerged in a fluid, the greater is the upthrust. - For the same volume of a body inside a fluid, the greater the density of the fluid, the greater is the upthrust. - The upthrust acts on the body in the upward direction, at the centre of buoyancy, that is, at the centre of gravity of the displaced fluid.
Each of these is worth looking at on its own.
5.2.1 Larger The Volume Submerged, Greater The Upthrust
You have already met this rule with your own hands. In the experiment of pushing a closed can into water, the upthrust grew steadily as more and more of the can went below the surface, and stopped growing only when the whole can was inside.
Notice carefully what the rule does not mention: the material of the body. Two bodies of the same size, submerged to the same extent in the same liquid, receive exactly the same upthrust, however different their materials may be. What the liquid responds to is the space the body takes up inside it, not what the body is made of.
A second everyday observation shows the same rule at work. Take a bunch of feathers and a pebble of the same mass and let them fall together in air. The pebble reaches the ground first. Because the feathers are far bulkier than the pebble for the same mass, they displace much more air, so the upthrust of the air on the feathers is much larger than the upthrust on the pebble, and it holds them back. Air resistance acts on the spread-out feathers as well and adds to the effect. Inside a vessel from which the air has been pumped out, there is no air to displace and no upthrust at all, so the feathers and the pebble fall together.
5.2.2 Denser The Fluid, Greater The Upthrust
Take two identical pieces of cork. Float one of them, cork A, in water, and the other, cork B, in glycerine. Both float, but you will see at once that the part of cork B inside the glycerine is smaller than the part of cork A inside the water. This is drawn in Fig. 5.2-1.
Diagram
Fig. 5.2-1 — Identical Corks Floating In Water And In Glycerine Two identical beakers side by side, each with a horizontal line for the liquid surface. Left beaker labelled ‘water’, with an identical cork block floating and a fairly large part of it shown below the surface, shaded and labelled ‘volume inside water’. Right beaker labelled ‘glycerine’, with the same-sized cork floating and a visibly smaller part below the surface, shaded and labelled ‘volume inside glycerine’. Label the corks ‘cork A’ and ‘cork B’.
Glycerine is denser than water. A floating body sinks only far enough to receive an upthrust equal to its own weight, and in glycerine the cork gets that much upthrust from a smaller submerged volume. If you now wanted to push cork B into the glycerine until it was as deep as cork A is in the water, you would have to press it down with an extra force. So, for the same volume of a body inside it, a denser liquid gives a greater upthrust.
5.2.3 Upthrust Acts At The Centre Of Buoyancy
A force does not only have a size and a direction; it also acts at a definite point. The weight of a body acts at its centre of gravity \(G\). The upthrust acts at a point called the centre of buoyancy.
For a uniform body completely immersed in a liquid, the displaced liquid fills exactly the space the body occupies, so the centre of buoyancy coincides with the centre of gravity of the body itself. This is the situation already drawn in Fig. 5.1-2, where both forces act along the same vertical line through the centre of the body.
For a body floating with only a part of it submerged, the two points are no longer the same, as Fig. 5.2-2 shows. Only the submerged part has displaced any liquid, so the centre of buoyancy \(B\) is at the centre of gravity of that submerged part, which lies below the centre of gravity \(G\) of the whole body. The weight \(W\) acts vertically downwards at \(G\), the upthrust \(F\_B\) acts vertically upwards at \(B\), and for the body to float,
\[W = F\_B\]
where \(W\) is the weight of the body in newton (N) and \(F\_B\) is the upthrust on its submerged part, also in newton (N).
Diagram
Fig. 5.2-2 — A Body Floating With Part Of It Submerged A vessel of liquid shown by horizontal dashes with a rectangular block floating so that roughly the lower half is below the liquid surface. Mark the point G at the centre of the whole block and the point B at the centre of the submerged part, directly below G. Draw an arrow vertically downwards from G labelled ‘W’ and an arrow vertically upwards from B labelled ‘F_B’. Mark the liquid surface line clearly and shade the submerged part.
5.3 Reason For Upthrust
We have seen what upthrust does. We have not yet asked where it comes from. Nothing pushes a body upward for no reason, and a liquid has no way of knowing whether a body has been dipped into it. The answer lies entirely in liquid pressure.
Recall two facts about a liquid held in a vessel. First, the liquid exerts pressure at every point in it and in every direction — upwards, downwards and sideways — and at any one point the pressure is the same in all these directions. Second, the pressure increases as you go deeper inside the liquid. Pressure is the thrust acting on unit area of a surface, and it is measured in pascal (Pa), where \(1\\ \\text{Pa} = 1\\ \\text{N m}^{-2}\).
Now consider a block of cross-sectional area \(A\) held immersed in a liquid, as drawn in Fig. 5.3-1. Its upper face and its lower face are not at the same depth. The lower face is deeper, so the pressure \(P\_2\) pushing upwards on it is more than the pressure \(P\_1\) pushing downwards on the upper face.
Diagram
Fig. 5.3-1 — A Block Immersed In A Liquid A vessel of liquid drawn with horizontal dashes, containing a rectangular block held below the surface. Draw a downward arrow onto the top face labelled ‘P_1’ and a longer upward arrow onto the bottom face labelled ‘P_2’. Draw a pair of equal horizontal arrows pressing inwards on the two vertical side faces, both labelled ‘F’, with a note ‘side thrusts are equal and opposite’. Mark the free liquid surface at the top and label the block’s cross-sectional area ‘A’.
So there is a difference of pressure, \(P\_2 - P\_1\), between the lower and the upper faces of the block. Since thrust is pressure multiplied by area, this pressure difference produces a net upward force on the block:
\[\\text{upthrust} = (P\_2 - P\_1)A\]
where \(P\_1\) and \(P\_2\) are the pressures on the upper and lower faces in pascal (Pa), \(A\) is the area of each face in \(\\text{m}^2\), and the upthrust is in newton (N). The relation as written applies to a block whose upper and lower faces are flat, horizontal and of equal area.
What about the side faces? Every point on one vertical side has a matching point at the same depth on the opposite side, and liquid pressure at the same depth is the same. So the sideways thrusts on the two sides are equal in size and opposite in direction, and they cancel out completely. Only the difference between the top and bottom thrusts is left over, and that difference is the upthrust.
This also explains a case that surprises many students. If a lamina, that is, a very thin sheet, is immersed in a liquid, its upper and lower surfaces are at almost the same depth. The pressures on the two surfaces are therefore almost the same, the liquid exerts a negligible upthrust on it, and the lamina sinks under its own weight.
5.4 Upthrust Is Equal To The Weight Of The Displaced Liquid
The pressure difference of the last section tells us why there is an upthrust. It can also tell us exactly how big that upthrust is. The result is remarkably simple and completely general:
When a body is immersed in a liquid, the upthrust on it due to the liquid is equal to the weight of the liquid displaced by the submerged part of the body.
5.4.1 Mathematical Proof
Consider a cylindrical body \(PQRS\) of cross-sectional area \(A\) immersed in a liquid of density \(\\rho\), as shown in Fig. 5.4-1. Let its upper surface \(PQ\) lie at a depth \(h\_1\) below the free surface of the liquid, and its lower surface \(RS\) at a greater depth \(h\_2\).
Diagram
Fig. 5.4-1 — Calculation Of The Upthrust On An Immersed Cylinder A vessel of liquid drawn with horizontal dashes, containing an upright cylinder held completely below the free surface. Label the top face corners P and Q and the bottom face corners S and R. From the free surface, draw a vertical dimension line down to the top face marked ‘h_1’ and a second vertical dimension line down to the bottom face marked ‘h_2’. Draw a downward arrow onto the top face labelled ‘F_1’ with ‘P_1’ beside it, and an upward arrow onto the bottom face labelled ‘F_2’ with ‘P_2’ beside it. Label the cross-sectional area of the cylinder ‘A’ and the liquid ‘liquid of density ρ’.
Thrust On The Two Faces
At depth \(h\_1\), the pressure on the upper surface \(PQ\) is
\[P\_1 = h\_1\\rho g\]
Since thrust is pressure multiplied by area, the downward thrust on the upper surface is
\[F\_1 = P\_1 \\times A = h\_1 \\rho g A \\qquad \\dots \\text{(i)}\]
At the greater depth \(h\_2\), the pressure on the lower surface \(RS\) is \(P\_2 = h\_2\\rho g\), so the upward thrust on the lower surface is
\[F\_2 = h\_2 \\rho g A \\qquad \\dots \\text{(ii)}\]
In both expressions, \(h\) is the depth in metre (m), \(\\rho\) is the density of the liquid in \(\\text{kg m}^{-3}\), \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\), \(A\) is the area in \(\\text{m}^2\), and the thrusts \(F\_1\) and \(F\_2\) are in newton (N).
The horizontal thrusts at the various points on the vertical sides of the cylinder balance one another, because liquid pressure is the same at all points at the same depth.
Finding The Resultant
Comparing equations (i) and (ii), \(F\_2 > F\_1\), because \(h\_2 > h\_1\). So the body experiences a net upward force. This resultant upward thrust is the buoyant force \(F\_B\):
\[F\_B = F\_2 - F\_1\] \[F\_B = h\_2\\rho g A - h\_1 \\rho g A\] \[F\_B = A(h\_2 - h\_1)\\rho g\]
Look at the quantity \(A(h\_2 - h\_1)\). Here \((h\_2 - h\_1)\) is the height of the cylinder and \(A\) is its cross-sectional area, so their product is simply the volume of the body inside the liquid. Writing that volume as \(V\),
\[F\_B = V\\rho g \\qquad \\dots (5.1)\]
where \(V\) is the volume of the body submerged in the liquid in \(\\text{m}^3\), \(\\rho\) is the density of the liquid in \(\\text{kg m}^{-3}\), \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\), and \(F\_B\) is the upthrust in newton (N).
Reading The Result As A Weight
The product \(V\\rho g\) can be read in stages, and each stage is a real physical quantity. A solid immersed in a liquid pushes aside a volume of liquid exactly equal to the volume of its own submerged part. So:
\[V\\rho g = \\text{volume of liquid displaced} \\times \\text{density of liquid} \\times g\]
But volume multiplied by density is mass, so
\[V\\rho g = \\text{mass of liquid displaced} \\times g = \\text{weight of the liquid displaced}\]
Hence
\[\\text{upthrust} = \\text{weight of the liquid displaced by the submerged part of the body} \\qquad \\dots (5.2)\]
The whole chain of reasoning, from unequal depths to a weight of liquid, is collected in FC 5.1.
Flowchart
FC 5.1 — From Pressure Difference To Weight Of Displaced Liquid Lower face lies deeper than upper face → pressure on lower face greater than on upper face → upward thrust F₂ greater than downward thrust F₁ → net upward force F_B = F₂ − F₁ = A(h₂ − h₁)ρg → F_B = Vρg, with V the submerged volume → V × density of liquid = mass of liquid displaced → F_B = weight of liquid displaced
Two points must be added to this result.
- If the body is completely immersed in the liquid, the volume of liquid displaced becomes equal to the body’s own full volume, and the upthrust then has its largest possible value \(F'\_B\).
- Although the result was derived for a cylinder, because that shape made the areas and depths easy to write down, it is equally true for a body of any shape and any size.
5.4.2 Factors Affecting The Upthrust
Equation (5.1) settles the question of what the upthrust depends on. Only two quantities appear in it, apart from \(g\):
- the volume of the body submerged in the liquid or fluid, and - the density of the liquid or fluid in which the body is submerged.
These are precisely the first two characteristic properties of upthrust met in section 5.2, now with a formula behind them.
5.4.3 Effect Of Upthrust
The effect of the upthrust is that the weight of a body immersed in a liquid appears to be less than its actual weight, and you met that effect in section 5.1 as apparent weight. What equation (5.2) now adds is a precise value for the loss: the apparent loss in weight is exactly the weight of the liquid displaced.
The effect is not confined to liquids. A body weighed on a sensitive spring balance in air weighs very slightly less than the same body weighed in a vacuum, because the air it displaces gives it a small upthrust. The difference is far too small for an ordinary balance to show, but a sensitive one records it.
5.5 Archimedes’ Principle
When a body is immersed in a liquid, it occupies space that the liquid was occupying a moment earlier. The liquid has to move out of the way, and we say the body has displaced the liquid. The volume of liquid displaced is equal to the volume of the submerged part of the body — no more and no less, because two things cannot be in the same place at the same time.
Section 5.4 showed that the upthrust on the body is equal in size to the weight of exactly that displaced liquid. It is this upthrust that makes an immersed body appear to weigh less than it really does, and the apparent loss in weight is equal to the upthrust on the body. Put together, these statements form Archimedes’ principle.
The principle is not restricted to liquids. It applies equally well to gases, since a body placed in a gas also displaces the gas around it and receives an upthrust equal to the weight of the gas displaced.
5.6 Experimental Verification Of Archimedes’ Principle
A principle in physics is only as good as the measurements that support it. Archimedes’ principle can be verified in the laboratory by either of the following two experiments. The first compares forces on a beam balance; the second measures a weight and a volume directly.
5.6.1 Verification Using A Solid And A Hollow Cylinder
The apparatus is shown in Fig. 5.6-1. Take two cylinders \(A\) and \(B\) of the same volume, \(A\) solid and \(B\) hollow. Suspend the two from the left arm of a physical balance, keeping the solid cylinder \(A\) hanging below the hollow cylinder \(B\). Balance the beam by placing weights on the right arm. At this stage both cylinders are in air.
Diagram
Fig. 5.6-1 — Verifying Archimedes’ Principle With A Solid And A Hollow Cylinder A physical balance with its beam horizontal. From the left arm hangs the hollow cylinder B, and directly below it, on the same thread, the solid cylinder A. Show A lowered into a beaker of water labelled D, standing on a bench labelled C, with A completely below the water surface and clear of the sides and bottom of the beaker. Show weights on the right pan. Label A ‘solid cylinder A’, B ‘hollow cylinder B (same volume as A)’, and add an arrow showing water being poured into the hollow cylinder B from a beaker.
Now immerse the solid cylinder \(A\) completely in water contained in a beaker \(D\) placed on a bench \(C\), taking care that the cylinder does not touch the sides or the bottom of the beaker. The beam no longer balances: the left arm rises, showing that cylinder \(A\) has lost weight. That loss is due to the upthrust of the water on cylinder \(A\).
Next, pour water gently into the hollow cylinder \(B\) until it is completely filled. The beam balances again.
What The Balancing Proves
The beam balanced in air, went out of balance by exactly the upthrust, and was brought back to balance by the water poured into \(B\). So the buoyant force on the solid cylinder \(A\) is equal to the weight of the water filling the hollow cylinder \(B\).
Now, \(A\) and \(B\) have equal volumes. Since \(A\) is completely immersed, the volume of water it displaces equals its own volume, which is also the volume of water needed to fill \(B\). Therefore the weight of water in \(B\) is just the weight of the water displaced by \(A\). The buoyant force on \(A\) is equal to the weight of water displaced by it, which is exactly what Archimedes’ principle claims.
5.6.2 Verification Using A Eureka Can
The second experiment measures the displaced water directly, using a eureka can — a vessel with a spout at the side, so that any liquid poured in above the level of the spout runs out of it. The arrangement is drawn in Fig. 5.6-2.
Take a solid, say a metallic piece, and suspend it by a thin thread from the hook of a spring balance. Note its weight in air. Now take a eureka can, fill it with water right up to its spout, and arrange a measuring cylinder below the spout.
Diagram
Fig. 5.6-2 — Verifying Archimedes’ Principle With A Eureka Can Two labelled parts side by side. Part (a): a spring balance with a graduated scale reading in gf, with a solid hanging from its hook by a thin thread, in air; label the reading ‘weight in air’. Part (b): the same spring balance with the solid now lowered fully into a eureka can filled to its spout; label the can ‘eureka can’, the side tube ‘spout’, and show water running from the spout into a measuring cylinder standing below; label the cylinder ‘measuring cylinder’ and mark a level on it as ‘100 ml’. Label the reading on the balance in part (b) ‘weight in water’.
Now immerse the solid gently into the water of the eureka can. The water displaced by it runs out of the spout and is collected in the measuring cylinder. When water stops dripping from the spout, note the weight of the solid as shown by the spring balance, and the volume of water collected in the measuring cylinder.
Working Out The Observations
Suppose the solid weighs \(300\\ \\text{gf}\) in air and \(200\\ \\text{gf}\) when it is completely immersed in water, and the volume of water collected in the measuring cylinder is \(100\\ \\text{ml}\), that is, \(100\\ \\text{cm}^3\). Then:
\[\\text{loss in weight} = 300\\ \\text{gf} - 200\\ \\text{gf} = 100\\ \\text{gf} \\qquad \\dots \\text{(i)}\]
Since the solid is completely immersed, the volume of water displaced is the volume of the solid, so
\[\\text{volume of water displaced} = 100\\ \\text{cm}^3\]
The density of water is \(1\\ \\text{g cm}^{-3}\), so \(100\\ \\text{cm}^3\) of water has a mass of \(100\\ \\text{g}\), and therefore
\[\\text{weight of water displaced} = 100\\ \\text{gf} \\qquad \\dots \\text{(ii)}\]
Comparing (i) and (ii),
\[\\text{weight of water displaced} = \\text{upthrust, or loss in weight}\]
Thus the weight of water displaced by a solid is equal to the loss in weight of the solid, and Archimedes’ principle is verified.
5.7 Why Denser Bodies Sink And Less Dense Bodies Float
So far, floating and sinking have been described by comparing two forces. There is a neater way to decide the question, using densities alone, and it follows directly from \(F\_B = V\\rho g\).
Let a body of volume \(V\) and density \(\\rho\) be immersed completely in a liquid of density \(\\rho\_L\). The weight of the body, acting downwards, is
\[W = V\\rho g\]
and the maximum upthrust on it, acting upwards, is
\[F'\_B = V\\rho\_L g\]
where \(V\) is in \(\\text{m}^3\), the densities are in \(\\text{kg m}^{-3}\), \(g\) is in \(\\text{m s}^{-2}\), and both forces are in newton (N). The same volume \(V\) appears in both, because the fully immersed body displaces liquid equal to its own volume. That is the key to the comparison: since \(V\) and \(g\) are common to both sides, comparing \(W\) with \(F'\_B\) is the same as comparing \(\\rho\) with \(\\rho\_L\).
The Three Cases
- If \(W > F'\_B\), that is \(V\\rho g > V\\rho\_L g\), that is \(\\rho > \\rho\_L\), the body sinks, pulled down by the unbalanced force \(W - F'\_B\).
- If \(W = F'\_B\), that is \(\\rho = \\rho\_L\), the body floats with the whole of it just inside the liquid, and the net force on it is zero.
- If \(W < F'\_B\), that is \(\\rho < \\rho\_L\), the body is pushed up by the unbalanced force \(F'\_B - W\) until only a part of it, of volume \(v\), remains inside the liquid — just enough for the upthrust \(F\_B = v\\rho\_L g\) on that part to balance the weight \(W\). The net force on the body is zero in this situation too.
Thus a body of density \(\\rho\) sinks in a liquid of density \(\\rho\_L\) if \(\\rho > \\rho\_L\), and floats if \(\\rho = \\rho\_L\) or \(\\rho < \\rho\_L\). Table 5.1 sets the three cases side by side.
Seeing It Happen
Experiment 1 — an iron nail and a piece of cork of the same mass. Place the iron nail on the surface of water in a cup. It sinks, which tells us that the weight of the nail pulling it down is greater than the upthrust of the water pushing it up. Now place the cork on the water surface. It floats. The upthrust on the cork, when fully immersed, would be more than its weight, because the density of water is more than the density of cork while the density of water is less than that of iron.
Diagram
Fig. 5.7-1 — An Iron Nail Sinks While A Cork Floats A single wide vessel of water drawn with horizontal dashes. On the left, an iron nail shown resting on the bottom of the vessel, labelled ‘iron nail (denser than water) — sinks’. On the right, a piece of cork floating at the surface with a small part below the surface line, labelled ‘cork (less dense than water) — floats’. Mark the free water surface as a clear horizontal line across the vessel.
Experiment 2 — solids of known density on water. Take a few solid bodies of different materials whose densities are known and place them gently on the surface of water. You will find that if the density of the material is equal to or less than the density of water, that is \(\\rho = \\rho\_w\) or \(\\rho < \\rho\_w\), the body floats, because the upthrust on its submerged part is equal to its own weight, \(F\_B = W\).
Different bodies float with different volumes inside water. If \(\\rho = \\rho\_w\), the body floats with the whole of its volume inside water. If \(\\rho < \\rho\_w\), it floats with only that much volume inside water for which the upthrust \(F\_B\) balances its weight \(W\). On the other hand, if the density of the material is more than the density of water, \(\\rho > \\rho\_w\), the body sinks, because the upthrust of the water on it is less than its weight, \(F\_B < W\).
Bodies That Are Not Solid Through And Through
An empty tin can, and equally an iron ship, floats on water even though the metal it is made of is far denser than water. What decides the matter is not the density of the material but the average density of the body — the mass of the body, counting both the material and the air enclosed inside it, divided by its total volume. A hollow body encloses a large volume of air, so its average density is small, and it floats.
5.8 Density
Everything in this chapter so far has turned on one quantity: how much matter a substance packs into a given space. It is time to define it properly.
If equal volumes of different substances such as wood, iron, zinc, water, glycerine and mercury are weighed on a physical balance, their masses turn out to be different. The mass of a piece of iron is much more than the mass of an equal volume of wood, because the particles of iron are heavier and more closely packed than those of wood. In other words, iron is denser than wood.
The same point can be made the other way round. Take equal masses of cotton and lead, say one kilogram of each, and their volumes are strikingly different. The volume of the cotton is much larger than the volume of an equal mass of lead, because the particles of lead are closely packed while those of cotton are very loose. Lead is denser than cotton.
To express both these facts — that equal volumes of different substances have different masses, and that equal masses of different substances have different volumes — we use the quantity called density.
\[\\text{density of a substance} = \\frac{\\text{mass of the substance}}{\\text{volume of the substance}} \\qquad \\dots (5.3)\]
Density is a scalar quantity, that is, it has magnitude but no direction, and it is represented by the letter \(\\rho\) (rho) or by \(d\). If the mass of a substance is \(M\) and its volume is \(V\), its density is
\[\\rho = \\frac{M}{V} \\qquad \\dots (5.4)\]
where \(M\) is the mass in kilogram (kg), \(V\) is the volume in \(\\text{m}^3\), and \(\\rho\) is the density in \(\\text{kg m}^{-3}\).
5.8.1 Unit Of Density
The unit of density follows straight from the definition:
\[\\text{unit of density} = \\frac{\\text{unit of mass}}{\\text{unit of volume}}\]
In the SI system, the unit of mass is the kilogram and the unit of volume is the cubic metre, so the SI unit of density is \(\\text{kg m}^{-3}\). In the C.G.S. system, the unit of mass is the gram and the unit of volume is the cubic centimetre, so the C.G.S. unit of density is \(\\text{g cm}^{-3}\), read as gram per cubic centimetre.
Relationship Between The Two Units
The two units are related by a fixed factor, and it is worth deriving it once rather than remembering it blindly:
\[1\\ \\text{kg m}^{-3} = \\frac{1\\ \\text{kg}}{1\\ \\text{m}^3} = \\frac{1000\\ \\text{g}}{(100\\ \\text{cm})^3} = \\frac{1000\\ \\text{g}}{10^6\\ \\text{cm}^3} = \\frac{1}{1000}\\ \\text{g cm}^{-3}\]
Thus
\[1\\ \\text{kg m}^{-3} = 10^{-3}\\ \\text{g cm}^{-3} \\qquad \\text{or} \\qquad 1\\ \\text{g cm}^{-3} = 1000\\ \\text{kg m}^{-3} \\qquad \\dots (5.5)\]
For example, the mass of \(1\\ \\text{cm}^3\) of iron is \(7.8\\ \\text{g}\), so the density of iron is \(7.8\\ \\text{g cm}^{-3}\), or \(7800\\ \\text{kg m}^{-3}\). Different substances have different densities, and this is one of the ways we tell materials apart.
5.8.2 Effect Of Temperature On Density
Most substances expand on heating and contract on cooling, while their mass remains unchanged. Since density is mass divided by volume, a larger volume for the same mass means a smaller density. Therefore the density of most substances decreases with an increase in temperature and increases with a decrease in temperature.
The Exception: Water
Water behaves very differently, because its expansion is uneven. When water is cooled from a high temperature, it contracts up to \(4^\\circ\\text{C}\); below \(4^\\circ\\text{C}\), and down to \(0^\\circ\\text{C}\), it expands instead. So the density of water gradually increases as it is cooled to \(4^\\circ\\text{C}\), and then starts decreasing as it is cooled further from \(4^\\circ\\text{C}\) to \(0^\\circ\\text{C}\). This behaviour is drawn in Fig. 5.8-1.
Diagram
Fig. 5.8-1 — Density Of Water Against Temperature A line graph. Horizontal axis labelled ‘temperature (°C)’ with marks at 0, 4, 10, 20. Vertical axis labelled ‘density of water (g cm⁻³)’ with a mark at 1.00 near the top. Draw a curve that rises from a point slightly below 1.00 at 0 °C to a peak exactly at 4 °C at the 1.00 line, then falls slowly and steadily as temperature increases to 20 °C. Mark the peak with a small dot and the label ‘maximum density at 4 °C’. Draw a light dashed vertical line from the peak down to the 4 on the temperature axis.
The peak of that curve is a standard value worth remembering:
\[\\text{density of water is maximum at } 4^\\circ\\text{C, equal to } 1\\ \\text{g cm}^{-3} \\text{ or } 1000\\ \\text{kg m}^{-3}\]
5.9 Relative Density
Densities are awkward numbers to carry about. It is often more useful to say how a substance compares with water than to quote its density in full. Since the density of water at \(4^\\circ\\text{C}\) is \(1\\ \\text{g cm}^{-3}\), or \(1000\\ \\text{kg m}^{-3}\), water makes a convenient standard, and the density of any substance can be compared with it. The ratio so obtained is called the relative density of that substance.
\[\\text{R.D.} = \\frac{\\text{density of substance } (\\rho\_s)}{\\text{density of water at } 4^\\circ\\text{C } (\\rho\_w)}\]
Since the density of a substance is the mass of unit volume of it, the same ratio can be written with masses in place of densities:
\[\\text{R.D.} = \\frac{\\text{mass of unit volume of substance}}{\\text{mass of unit volume of water at } 4^\\circ\\text{C}} = \\frac{\\text{mass of a certain volume of a substance }(m\_s)}{\\text{mass of an equal volume of water at } 4^\\circ\\text{C }(m\_w)} \\qquad \\dots (5.6)\]
The second form only works if the two volumes compared are equal, which is why the phrase ‘an equal volume of water’ cannot be dropped from it.
Because relative density is a pure ratio, the units of the two densities cancel each other, provided both are expressed in the same unit. So relative density has no unit at all. Like density, it is a scalar quantity.
5.9.1 Relationship Between Density And Relative Density
Since relative density is worked out by dividing by the density of water, its numerical value is tied very simply to the numerical value of the density — but the tie depends on which system of units you started in.
In The C.G.S. System
The density of water at \(4^\\circ\\text{C}\) is \(1\\ \\text{g cm}^{-3}\), so
\[\\text{R.D.} = \\frac{\\text{density of substance in g cm}^{-3}}{1.0\\ \\text{g cm}^{-3}} \\qquad \\dots (5.7)\]
or, in words, R.D. \(=\) numerical value of the density expressed in \(\\text{g cm}^{-3}\). Dividing by one changes nothing except the unit, which disappears.
In The SI System
The density of water at \(4^\\circ\\text{C}\) is \(1000\\ \\text{kg m}^{-3}\), so
\[\\text{R.D.} = \\frac{\\text{density of substance in kg m}^{-3}}{1000\\ \\text{kg m}^{-3}} \\qquad \\dots (5.8)\]
or R.D. \(=\) numerical value of the density in \(\\text{kg m}^{-3}\), multiplied by \(10^{-3}\).
Three examples show how quickly this can be used:
- The density of copper is \(8.9\\ \\text{g cm}^{-3}\), so its R.D. is \(8.9\). - The density of mercury is \(13.6 \\times 10^{3}\\ \\text{kg m}^{-3}\), so its R.D. is \(13.6\). - The R.D. of silver is \(10.8\), so its density is \(10.8\\ \\text{g cm}^{-3}\) in C.G.S. units and \(10.8 \\times 10^{3}\\ \\text{kg m}^{-3}\) in SI units.
The two quantities are compared point by point in Table 5.2.
Table 5.3 gives the density and relative density of some common substances. Read it as a reference table rather than a list to be memorised, but do notice which substances lie below the density of water and which lie above, because that single comparison decides whether they float in it.
Table 5.3 — Density and relative density of some common substances
| Substance | Density (\(\\text{kg m}^{-3}\)) | Density (\(\\text{g cm}^{-3}\)) | Relative density |
|---|---|---|---|
| Cork | 240 | 0.24 | 0.24 |
| Wood (pine) | 500 | 0.50 | 0.50 |
| Petrol | 800 | 0.80 | 0.80 |
| Turpentine | 870 | 0.87 | 0.87 |
| Ice | 920 | 0.92 | 0.92 |
| Olive oil | 920 | 0.92 | 0.92 |
| Pure water (at \(4^\\circ\\text{C}\)) | 1000 | 1.00 | 1 |
| Sea water | 1025 | 1.02 | 1.02 |
| Glycerine | 1260 | 1.26 | 1.26 |
| Glass | 2500 | 2.5 | 2.5 |
| Aluminium | 2700 | 2.70 | 2.70 |
| Iron | 7860 | 7.86 | 7.86 |
| Copper | 8920 | 8.92 | 8.92 |
| Silver | 10500 | 10.5 | 10.5 |
| Mercury | 13600 | 13.6 | 13.6 |
| Gold | 19300 | 19.3 | 19.3 |
| Platinum | 21500 | 21.5 | 21.5 |
5.10 Determination Of Relative Density Of A Solid By Archimedes’ Principle
Relative density was defined as a comparison of masses:
\[\\text{R.D.} = \\frac{\\text{mass of the body}}{\\text{mass of water (at } 4^\\circ\\text{C) of volume equal to that of the body}}\]
The trouble is the denominator. How do you obtain a volume of water exactly equal to the volume of an irregular solid? Archimedes’ principle answers this at once. A body completely immersed in water displaces water equal to its own volume, so the water displaced is the equal volume of water we need. Its mass can be found from the loss in weight of the body.
Working the ratio down step by step,
\[\\text{R.D.} = \\frac{\\text{mass of body}}{\\text{mass of water displaced by the body}} = \\frac{\\text{weight of body}}{\\text{weight of water displaced by the body}}\]
and since the weight of water displaced equals the upthrust, which equals the loss in weight of the body in water,
\[\\text{R.D.} = \\frac{\\text{weight of body}}{\\text{loss in weight of the body in water}}\]
\[\\text{R.D.} = \\frac{\\text{weight of body in air}}{\\text{weight of body in air} - \\text{weight of body in water}} \\qquad \\dots (5.9)\]
Going from masses to weights in the second step is allowed because weight is mass multiplied by \(g\), and the same \(g\) appears in the numerator and the denominator, so it cancels. If the weight of the body in air is \(W\_1\) and its weight in water is \(W\_2\), then
\[\\text{R.D.} = \\frac{W\_1}{W\_1 - W\_2} \\qquad \\dots (5.10)\]
where \(W\_1\) and \(W\_2\) are both weights, measured in the same unit — gf, kgf or newton. Since R.D. is a ratio, the unit cancels, and the answer is a pure number.
Note carefully how mass and weight are quoted here. Weight and mass are related as weight \(=\) mass \(\\times\) acceleration due to gravity, that is \(W = Mg\). On weighing a body with a physical balance, its mass is expressed in kg or g, while its weight is expressed in kgf or gf.
We now describe the procedure for two cases: when the solid is denser than water and insoluble in it, and when the solid is denser than water but soluble in it.
5.10.1 Relative Density Of A Solid Denser Than Water And Insoluble In It
Procedure
- Suspend a piece of the given solid with a thread from the hook of the left pan of a physical balance, and find its weight \(W\_1\).
- Now place a wooden bridge over the left pan of the balance and place a beaker nearly two-thirds filled with water on the bridge. Take care that the bridge and the beaker do not touch the pan of the balance. The bridge carries the beaker’s weight down to the bench instead of onto the pan, so the balance goes on weighing only the solid.
- Immerse the solid completely in the water so that it does not touch the walls or the bottom of the beaker, as shown in Fig. 5.10-1, and find the weight \(W\_2\) of the solid in water. If the solid touched the glass, the beaker would take part of its weight and the reading would be wrong.
Diagram
Fig. 5.10-1 — Finding The Relative Density Of A Solid Denser Than Water A physical balance with its left pan shown. A wooden bridge straddles the left pan without touching it, standing on the base board; label it ‘wooden bridge’. A beaker about two-thirds filled with water rests on the bridge; label ‘water’. A solid hangs from the hook under the left pan on a thread and is completely below the water surface, clear of the walls and the bottom; label it ‘solid’. Show the beam and the right pan of the balance, and mark that the thread passes into the beaker without touching its rim.
Observations And Calculations
Weight of solid in air \(= W\_1\) gf Weight of solid in water \(= W\_2\) gf
Loss in weight of the solid when immersed in water \(= (W\_1 - W\_2)\) gf
\[\\text{R.D.} = \\frac{\\text{weight of solid in air}}{\\text{loss in weight of solid in water}}\]
\[\\text{R.D.} = \\frac{W\_1}{W\_1 - W\_2} \\qquad \\dots (5.11)\]
The whole experiment is summarised in FC 5.2.
Flowchart
FC 5.2 — Steps In Finding The Relative Density Of An Insoluble Solid Suspend the solid from the hook of the left pan and note its weight in air, W₁ → place the wooden bridge over the pan, without touching it → place a beaker two-thirds filled with water on the bridge → lower the solid until it is completely immersed, clear of the walls and bottom → note its weight in water, W₂ → find the loss in weight, W₁ − W₂ → calculate R.D. = W₁ / (W₁ − W₂)
5.10.2 Relative Density Of A Solid Denser Than Water And Soluble In It
Some solids dissolve in water, and for these the experiment above would destroy the very sample being weighed. The way round it is to change the liquid.
If the solid is soluble in water, then instead of water we take a liquid of known relative density in which the solid is insoluble, and in which it sinks. The process described above is then repeated with that liquid in the beaker. Since the loss in weight now gives the weight of an equal volume of the liquid, not of water, the result must be scaled by the relative density of that liquid:
\[\\text{R.D. of solid} = \\frac{\\text{weight of solid in air}}{\\text{loss in weight of solid in liquid}} \\times \\text{R.D. of liquid} \\qquad \\dots (5.12)\]
5.11 Determination Of Relative Density Of A Liquid By Archimedes’ Principle
The relative density of a liquid can be found by the same method, with one extra weighing. By definition,
\[\\text{R.D. of a liquid} = \\frac{\\text{weight of a given volume of the liquid}}{\\text{weight of the same volume of water}} \\qquad \\dots (5.13)\]
Once again the difficulty is in getting the two volumes exactly equal, and once again Archimedes’ principle removes it. If one and the same solid is immersed first in the liquid and then in water, it displaces a volume of each equal to its own volume — so the two displaced volumes are automatically equal. Equation (5.13) therefore takes the form
\[\\text{R.D.} = \\frac{\\text{weight of liquid displaced by a body}}{\\text{weight of water displaced by the same body}}\]
\[\\text{R.D.} = \\frac{\\text{weight of body in air} - \\text{weight of body in liquid}}{\\text{weight of body in air} - \\text{weight of body in water}} \\qquad \\dots (5.14)\]
The second form follows from the first because each loss in weight is, by Archimedes’ principle, the weight of the fluid displaced.
The Body To Be Used
The choice of the solid is not free. It must be heavier than both the given liquid and water, so that it sinks completely in each of them, and it must be insoluble in both, so that it is the same body at every weighing.
The body is first weighed in air, then in the liquid, and then — after being washed with water and dried, so that no trace of the liquid clings to it — it is weighed in water. The three weighings are shown in Fig. 5.11-1.
Diagram
Fig. 5.11-1 — Three Weighings Used To Find The Relative Density Of A Liquid Three sketches side by side, each showing the same metal solid hanging by a thread from the hook of a spring balance with a graduated scale in gf. Sketch (a): the solid hanging in air, labelled ‘weight in air W₁’. Sketch (b): the solid completely immersed in a beaker of the given liquid, clear of walls and bottom, labelled ‘weight in liquid W₂’ and the beaker labelled ‘given liquid’. Sketch (c): the solid completely immersed in a beaker of water, labelled ‘weight in water W₃’ and the beaker labelled ‘water’. Add a note under sketch (c): ‘solid washed and dried before this weighing’.
If the weight of the body in air is \(W\_1\) gf, in the liquid is \(W\_2\) gf and in water is \(W\_3\) gf, then from equation (5.14),
\[\\text{R.D. of liquid} = \\frac{W\_1 - W\_2}{W\_1 - W\_3} \\qquad \\dots (5.15)\]
where all three weights are measured in the same unit, so that the unit cancels and the relative density comes out as a pure number.
5.12 Principle Of Floatation
We now have everything needed to state the rule that governs every floating object, from a cork to a cargo ship.
When a body is immersed in a liquid, two forces act on it.
- The weight \(W\) of the body, acting vertically downwards through its centre of gravity \(G\). This force tends to sink the body.
- The upthrust \(F\_B\) of the liquid, acting vertically upwards through the centre of buoyancy \(B\), that is, the centre of gravity of the displaced liquid. The upthrust is equal in magnitude to the weight of the liquid displaced, and it tends to make the body float.
These are the two forces already drawn for a floating body in Fig. 5.2-2. Written out in full,
\[W = \\text{volume of body} \\times \\text{density of body} \\times g \\qquad \\dots (5.16)\]
\[F\_B = \\text{volume of submerged part of body} \\times \\text{density of liquid} \\times g \\qquad \\dots (5.17)\]
where the volumes are in \(\\text{m}^3\), the densities in \(\\text{kg m}^{-3}\), \(g\) is in \(\\text{m s}^{-2}\), and both forces are in newton (N). The difference between the two lines is the volume used: the whole volume for the weight, but only the submerged volume for the upthrust. The upthrust reaches its maximum value \(F'\_B\) when the body is completely immersed inside the liquid, for then the two volumes become the same.
5.12.1 The Three Cases Of Sinking And Floating
Whether the body sinks or floats depends on whether that maximum upthrust \(F'\_B\) is less than, equal to, or greater than the weight \(W\). The three outcomes are drawn together in Fig. 5.12-1.
Diagram
Fig. 5.12-1 — The Three Cases Of Sinking And Floating Three panels side by side, each a vessel of liquid drawn with horizontal dashes and a small square block. Panel (a): the block resting at the bottom of the vessel, with an upward arrow labelled ‘F′_B’ shorter than the downward arrow labelled ‘W’; caption line ‘W > F′_B: the body sinks’. Panel (b): the block just below the free surface with its whole volume immersed, with the upward arrow ‘F′_B’ and downward arrow ‘W’ drawn equal in length; caption line ‘W = F′_B: floats fully immersed’. Panel (c): the block floating with part of it above the surface, marked with G at the centre of the block and B at the centre of the submerged part below G, an upward arrow ‘F_B’ from B and a downward arrow ‘W’ from G, equal in length; caption line ‘W < F′_B: floats partly immersed’.
Case (i): \(W > F'\_B\), that is, the weight of the body is greater than the weight of the displaced liquid. The body sinks. If a spring balance were attached to it, the apparent weight it would read is \((W - F'\_B)\), acting vertically downwards. This is the case when the density \(\\rho\) of the solid is greater than the density \(\\rho\_L\) of the liquid.
Case (ii): \(W = F'\_B\), that is, the weight of the body is equal to the weight of the displaced liquid. The body floats just below the surface of the liquid, with the whole of it immersed. Its apparent weight is zero, and its density \(\\rho\) is equal to the density \(\\rho\_L\) of the liquid.
Case (iii): \(W < F'\_B\), that is, the weight of the body is less than the weight of the liquid it would displace on full immersion. The body floats partly above and partly below the surface. Only that much portion of it gets submerged for which the weight of the displaced liquid becomes equal to the weight of the body. While it floats, the weight \(W\) acts at the centre of gravity \(G\), the upthrust \(F\_B\) acts at the centre of buoyancy \(B\) which is vertically below \(G\), and \(W\) is balanced by the upthrust due to the submerged part alone. The apparent weight of the body is zero in this case too, and its density \(\\rho\) is less than the density \(\\rho\_L\) of the liquid.
What Cases (ii) And (iii) Have In Common
In both of the floating cases, whatever fraction of the body lies under the surface, the same statement holds:
\[\\text{weight of body} = \\text{weight of liquid displaced by the submerged part of the body}\]
\[W = F\_B \\qquad \\dots (5.18)\]
and therefore the apparent weight of a floating body is zero. This result is the principle of floatation.
A body floating in different liquids floats with different extents of its volume submerged, but in each case the weight of the liquid displaced — that is, the buoyant force on the body due to that liquid — is equal to the weight of the body. The body’s weight has not changed, so neither can the weight of liquid it must displace.
5.13 Relation Between The Submerged Volume Of A Floating Body And The Two Densities
The principle of floatation tells us that a floating body displaces its own weight of liquid. It does not yet tell us how much of the body will be under the surface. That is what this section settles, and the answer turns out to depend on nothing but the two densities.
Let \(V\) be the total volume of a body of density \(\\rho\_s\), floating with a volume \(v\) immersed inside a liquid of density \(\\rho\_L\), as drawn in Fig. 5.13-1.
Diagram
Fig. 5.13-1 — A Floating Body: Total Volume And Submerged Volume A vessel of liquid drawn with horizontal dashes, with a rectangular block floating so that roughly two-thirds of it lies below the free surface. Shade the submerged part and label it ‘v = submerged volume’. Bracket the whole block at one side and label it ‘V = total volume’. Label the block ‘body of density ρ_s’ and the liquid ‘liquid of density ρ_L’. Draw an upward arrow from the submerged part labelled ‘F_B’ and a downward arrow from the centre of the block labelled ‘W’.
The weight of the body uses its whole volume:
\[W = \\text{volume of body} \\times \\text{density of body} \\times g = V\\rho\_s g\]
The upthrust, which is the weight of liquid displaced by the immersed part, uses only the submerged volume:
\[F\_B = \\text{volume of displaced liquid} \\times \\text{density of liquid} \\times g = v\\rho\_L g\]
For floatation, \(W = F\_B\), that is,
\[V\\rho\_s g = v\\rho\_L g\]
The acceleration due to gravity \(g\) appears on both sides and cancels. Dividing both sides by \(V\\rho\_L\) then leaves
\[\\frac{v}{V} = \\frac{\\rho\_s}{\\rho\_L} \\qquad \\dots (5.19)\]
where \(v\) is the submerged volume and \(V\) the total volume, in the same unit of volume, and \(\\rho\_s\) and \(\\rho\_L\) are the densities of the body and of the liquid, in the same unit of density. Both sides are pure ratios with no unit. The relation holds only for a body floating freely, and only while \(\\rho\_s \\leq \\rho\_L\) — if the body’s density were greater, the ratio would come out more than one, which is impossible, and that is the algebra’s way of saying the body sinks.
In words,
\[\\frac{\\text{volume of immersed part of body}}{\\text{total volume of body}} = \\frac{\\text{density of body}}{\\text{density of liquid}}\]
Two Examples Of The Relation
- A cork of density \(\\rho\_s = \\tfrac{3}{4}\\ \\text{g cm}^{-3}\) floating in water, for which \(\\rho\_L = 1\\ \\text{g cm}^{-3}\), will have \(\\tfrac{3}{4}\) of its volume immersed inside the water and \(\\tfrac{1}{4}\) of its volume outside the surface of the water.
- A cube of ice of density \(\\rho\_s = 0.9\\ \\text{g cm}^{-3}\) floating in water of density \(\\rho\_L = 1\\ \\text{g cm}^{-3}\) will have \(90%\) of its volume immersed in the water, with only \(10%\) standing outside the surface.
Notice how the fraction outside is found: it is whatever is left over after subtracting the submerged fraction from the whole.
What The Relation Tells Us
Inference 1. Of two floating bodies of the same volume, the one of higher density has more of its volume submerged inside the liquid, and so displaces more liquid and experiences a higher upthrust than the body of lower density. That larger upthrust is, of course, still exactly equal to its own larger weight.
Inference 2. A body floats in liquids of different densities only when the density of the body is less than or equal to the density of the liquid. The volume of the body submerged in the liquid decreases as the density of the liquid increases, because a denser liquid supplies the same upthrust from a smaller displaced volume.
In each of these cases, the upthrust given by the liquid — that is, the weight of liquid displaced by the submerged part — is the same, and is equal to the weight of the body. Only the volume needed to produce it changes.
5.14 Applications Of The Principle Of Floatation
One short relation, \(W = F\_B\), explains a surprising range of things: steel ships, swimmers, submarines, icebergs, fish and balloons. Each of the following is the same physics wearing different clothes.
5.14.1 Floatation Of An Iron Ship
Section 5.7 explained why an iron nail sinks in water while a hollow iron ship floats on it: the nail has the density of iron, while the ship, with its large enclosed volume of air, has an average density less than that of water. Even with only a small portion of the ship submerged, the weight of water displaced by that portion becomes equal to the total weight of the ship, and so it floats. What remains to be explained is how a ship behaves as its load and its waters change.
Loading And Unloading
When cargo is loaded onto a sailing ship, its weight increases. It must therefore displace more water, so it sinks further into the water until the weight of water displaced by its submerged part becomes equal to the weight of the loaded ship. If cargo is unloaded, the ship rises in the water until the weight of water displaced once again balances the weight of the unloaded ship. A loaded ship is submerged more; an unloaded ship is submerged less.
From Sea Water To River Water
River water is of lower density than sea water, and even the water of different seas differs in density. So when a ship sails from a sea of higher water density into water of lower density — from sea water into river water, for instance — it sinks further. By the law of floatation, the ship must still displace a weight of water equal to its own weight, and in water of lower density a greater volume must be displaced to give that same weight.
The Plimsoll Line
Each ship has a white line painted on its side, known as the Plimsoll line. This line indicates the safe limit for loading the ship in water of density \(10^{3}\\ \\text{kg m}^{-3}\). A ship is not allowed to be loaded further once its Plimsoll line starts touching the water level, so that when it sails in sea water of density more than \(10^{3}\\ \\text{kg m}^{-3}\), only the part of it below the Plimsoll line remains submerged.
Ballast In An Unloaded Ship
An unloaded ship floats with a very small volume inside water. As a result its centre of gravity is higher, and its equilibrium is unstable — there is a danger of it being blown over on its side by strong winds. An unloaded ship is therefore filled with sand or stones, called ballast, at its bottom. This lowers its centre of gravity and makes its equilibrium stable.
5.14.2 Floatation Of The Human Body
The average density of the human body depends on the proportion of its constituents such as bone, blood, muscle and fat, since each of these has a different density. It also depends on how much air is in the lungs at the time. The average density of the body with empty lungs is \(1.07\\ \\text{g cm}^{-3}\), while with the lungs filled with air it is \(1.00\\ \\text{g cm}^{-3}\).
A good swimmer can float on water like any floating object, with the lungs filled with air and the nose and mouth projecting just above the water surface. The weight of water displaced is then nearly equal to the swimmer’s own weight, so very little effort is needed to stay afloat.
Sea Water, River Water And The Dead Sea
It is easier for a person to swim in sea water than in fresh or river water. Because of the minerals dissolved in it, the density of sea water, about \(1.026\\ \\text{g cm}^{-3}\), is more than the density of fresh water, \(1.0\\ \\text{g cm}^{-3}\). In sea water a smaller portion of the body needs to be submerged before the weight of water displaced equals the whole weight of the body, whereas in fresh water a larger portion must go under to displace the same weight. In the Dead Sea the water is much denser still, about \(1.16\\ \\text{g cm}^{-3}\), and a person can float there with only a small portion of the body submerged.
5.14.3 Floatation Of Submarines
A submarine is a fish-shaped, water-tight boat provided with several ballast, or floatation, tanks in its front and rear parts. It is fitted with periscopes so that the crew can see above the water surface even when the submarine is well inside the water. Fig. 5.14-1 shows the arrangement.
Diagram
Fig. 5.14-1 — Ballast Tanks Of A Submarine Side view of a submarine below the water surface in a sea drawn with horizontal dashes. Show the hull in outline with two shaded compartments, one near the front and one near the rear, both labelled ‘ballast tank’. Show a periscope tube rising from the top of the hull to just above the water surface, labelled ‘periscope’. Mark the sea surface as a clear horizontal line above the submarine.
A submarine can be made to dive into water or to rise up to the surface as and when desired. If the submarine is to dive, its ballast tanks are filled with water, so that the average density of the submarine becomes greater than the density of sea water, and it sinks. If it is to rise, the water is forced out of the ballast tanks by letting compressed air into them. This makes the average density of the submarine less than that of sea water, so the weight of water displaced by its partly submerged volume becomes equal to the weight of the submarine and it rises to the surface. The two routes are set out in FC 5.3.
Flowchart
FC 5.3 — Diving And Surfacing Of A Submarine To dive: ballast tanks filled with sea water → mass of submarine increases while its volume stays the same → average density becomes greater than that of sea water → weight exceeds the maximum upthrust → submarine sinks To rise: compressed air forced into ballast tanks → water driven out of the tanks → average density becomes less than that of sea water → upthrust exceeds weight → submarine rises to the surface and floats partly submerged
5.14.4 Floatation Of Icebergs
The density of ice, \(0.917\\ \\text{g cm}^{-3}\), is less than the density of water, \(1\\ \\text{g cm}^{-3}\). Huge masses of floating ice, called icebergs, are therefore able to float on water with their major part inside the water surface and only a small portion above it.
Volume Of An Iceberg Above The Water Surface
If the total volume of an iceberg is \(V\) and the volume submerged is \(v\), then by the principle of floatation,
\[\\text{weight of water displaced by the submerged part} = \\text{total weight of the iceberg}\]
\[v \\times \\rho\_{\\text{water}} \\times g = V \\times \\rho\_{\\text{ice}} \\times g\]
\[\\frac{v}{V} = \\frac{\\rho\_{\\text{ice}}}{\\rho\_{\\text{water}}} \\qquad \\dots (5.20)\]
where \(v\) and \(V\) are volumes in the same unit and the two densities are in the same unit, so the ratio is a pure number.
- An iceberg of density \(\\rho\_{\\text{ice}} = 0.917\\ \\text{g cm}^{-3}\) floating on fresh water of density \(\\rho\_{\\text{water}} = 1.0\\ \\text{g cm}^{-3}\) has \(v = 0.917,V\), that is, \(91.7%\) of its total volume below the water surface, and only \(8.3%\) of its volume above the surface, as shown in Fig. 5.14-2.
- The same iceberg floating on sea water of density \(\\rho\_{\\text{water}} = 1.026\\ \\text{g cm}^{-3}\) has \(v = \\dfrac{0.917}{1.026}V = 0.893,V\), that is, \(89.3%\) of its total volume submerged inside sea water and only \(10.7%\) above the sea water surface.
Diagram
Fig. 5.14-2 — A Floating Iceberg A sea drawn with horizontal dashes and a clear surface line. A large irregular mass of ice floats with only a small tip above the surface line and the bulk of it, drawn much larger, below the surface; label the tip ‘about one-tenth above the surface’ and the lower mass ‘the major part lies below the surface’. To one side, on the surface, draw a small ship approaching, labelled ‘ship’. Add an arrow from the ship towards the iceberg.
Why Icebergs Are Dangerous For Ships
Icebergs float with their major part, nearly \(90%\), inside the water and only a small part outside. Since the portion of an iceberg inside the water depends on the density of the sea water, it becomes difficult for those steering a ship to judge the true size of the iceberg from the little that shows. An iceberg is therefore very dangerous for a ship, which may collide with it and be damaged.
No Change In Water Level When Floating Ice Melts
When a floating piece of ice melts, the water formed has exactly the volume of the part of the ice that was below the surface, because that is the volume of water whose weight was supporting the ice. The melted water therefore fills precisely the space the ice occupied below the surface, and the level of the water does not change.
5.14.5 Floatation Of Fish
Many species of fish have an organ called a swim bladder, which acts like the ballast or floatation tank of a submarine.
When a fish has to rise up in water, it diffuses oxygen gas from its blood into the bladder. The volume of the bladder increases and the average density of the fish decreases. This increases the volume of water displaced by the fish, so the upthrust on it increases and it rises. When the fish has to come down, it empties its bladder to the required extent, so its volume decreases and its density increases. The upthrust on the fish then decreases and it goes down in the water.
5.14.6 Rising Of Balloons
When a light gas such as hydrogen or helium, whose density is much less than that of air, is filled into a balloon, the weight of the air displaced by the inflated balloon — that is, the upthrust — becomes more than the weight of the gas-filled balloon, and the balloon rises up. Modern balloons are usually filled with helium rather than hydrogen, because hydrogen is highly flammable while helium does not burn at all.
Why A Balloon Does Not Rise Indefinitely
The density of air decreases with altitude. So as the balloon gradually goes up, the weight of the displaced air, and hence the upthrust on it, decreases. The balloon keeps on rising as long as the upthrust on it exceeds its weight. When the upthrust becomes equal to its weight, it stops rising further and floats at that height.
5.15 Summary
- When a body is partly or wholly immersed in a fluid, the fluid exerts an upward force on it, called the upthrust or buoyant force, \(F\_B\). Its SI unit is the newton (N); it is also expressed in kgf or gf.
- The property of a fluid by which it exerts this upward force on a body placed in it is called buoyancy.
- Because the upthrust opposes the weight, a body in a fluid seems lighter than it really is: \(\\text{apparent weight} = \\text{real weight} - \\text{upthrust}\), all three measured in newton (N). The mass of the body does not change.
- The upthrust has three characteristic properties: the larger the volume of the body submerged, the greater the upthrust; for the same submerged volume, the denser the fluid, the greater the upthrust; and the upthrust acts vertically upwards at the centre of buoyancy, which is the centre of gravity of the displaced fluid.
- Upthrust arises because liquid pressure increases with depth. The lower face of an immersed body is under greater pressure than its upper face, while the sideways thrusts cancel, leaving a net upward force \(\\text{upthrust} = (P\_2 - P\_1)A\), where \(P\_1\) and \(P\_2\) are the pressures on the upper and lower faces in pascal (Pa) and \(A\) is the area of each face in \(\\text{m}^2\).
- For a body immersed in a liquid of density \(\\rho\), \[F\_B = V\\rho g\] where \(V\) is the volume of the body submerged in \(\\text{m}^3\), \(\\rho\) is the density of the liquid in \(\\text{kg m}^{-3}\), \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\), and \(F\_B\) is the upthrust in newton (N).
- Since \(V\\rho g\) is the weight of the displaced liquid, the upthrust is equal to the weight of the liquid displaced by the submerged part of the body. The result was derived for a cylinder but holds for a body of any shape and size.
- The upthrust on a body depends on only two things: the volume of the body submerged, and the density of the fluid.
- Archimedes’ principle states that 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. It applies to gases as well as to liquids.
- The principle is verified in the laboratory either by balancing the loss in weight of a solid cylinder against the weight of water filling an identical hollow cylinder, or by collecting the water displaced from a eureka can and showing that its weight equals the loss in weight of the solid.
- For a body of density \(\\rho\) fully immersed in a liquid of density \(\\rho\_L\): it sinks if \(\\rho > \\rho\_L\); it floats fully immersed if \(\\rho = \\rho\_L\); and it floats partly immersed if \(\\rho < \\rho\_L\). A hollow body floats because its average density, which counts the enclosed air, is less than that of the liquid.
- Density is mass per unit volume: \[\\rho = \\frac{M}{V}\] where \(M\) is the mass in kilogram (kg), \(V\) the volume in \(\\text{m}^3\) and \(\\rho\) the density in \(\\text{kg m}^{-3}\). In C.G.S. units it is in \(\\text{g cm}^{-3}\), and \(1\\ \\text{g cm}^{-3} = 1000\\ \\text{kg m}^{-3}\). Density is a scalar.
- The density of most substances decreases with a rise in temperature and increases with a fall, since the mass stays the same while the volume changes. Water is an exception: its density is maximum at \(4^\\circ\\text{C}\), equal to \(1\\ \\text{g cm}^{-3}\) or \(1000\\ \\text{kg m}^{-3}\).
- The relative density of a substance is the ratio of its density to the density of water at \(4^\\circ\\text{C}\), or equally the ratio of the mass of a certain volume of the substance to the mass of an equal volume of water at \(4^\\circ\\text{C}\). It is a scalar and has no unit.
- For a solid denser than water and insoluble in it, \[\\text{R.D.} = \\frac{W\_1}{W\_1 - W\_2}\] where \(W\_1\) is the weight of the solid in air and \(W\_2\) its weight in water, both in the same unit. If the solid is soluble in water, a liquid of known relative density is used instead, and \(\\text{R.D. of solid} = \\dfrac{\\text{weight in air}}{\\text{loss in weight in liquid}} \\times \\text{R.D. of liquid}\).
- For a liquid, using one body weighed in air, in the liquid and in water, \[\\text{R.D. of liquid} = \\frac{W\_1 - W\_2}{W\_1 - W\_3}\] where \(W\_1\), \(W\_2\) and \(W\_3\) are the weights of the body in air, in the liquid and in water, all in the same unit.
- Principle of floatation: the weight of a floating body is equal to the weight of the liquid displaced by its submerged part, that is, \(W = F\_B\). The apparent weight of a floating body is zero.
- For a freely floating body, \[\\frac{v}{V} = \\frac{\\rho\_s}{\\rho\_L}\] where \(v\) is the submerged volume and \(V\) the total volume in the same unit, \(\\rho\_s\) is the density of the body and \(\\rho\_L\) that of the liquid in the same unit. The relation holds only while \(\\rho\_s \\leq \\rho\_L\).
- A ship floats because its average density is less than that of water; it sinks deeper when loaded and when it sails from sea water into river water, and the Plimsoll line marks its safe loading limit. Ballast at the bottom of an unloaded ship lowers its centre of gravity and makes its equilibrium stable.
- A swimmer floats more easily in sea water than in river water because sea water is denser. A submarine dives by filling its ballast tanks with water and rises by forcing the water out with compressed air. A fish rises and sinks by changing the volume of its swim bladder.
- An iceberg floats with about \(91.7%\) of its volume below the surface in fresh water and about \(89.3%\) below the surface in sea water, which is why so little of it is visible and why it is dangerous to ships. When floating ice melts, the water level does not change.
- A balloon filled with a gas less dense than air rises because the weight of air it displaces exceeds its own weight. It stops rising at the height where the upthrust, falling as the air thins, becomes equal to its weight.
Flowchart
FC 5.4 — The Spine Of The Chapter Liquid pressure increases with depth → greater thrust on the lower face than on the upper face → net upward force, the upthrust → upthrust F_B = Vρg = weight of liquid displaced (Archimedes’ principle) → compare the maximum upthrust with the weight → ρ > ρ_L: body sinks; ρ ≤ ρ_L: body floats → for a floating body, W = F_B (principle of floatation) → submerged fraction v/V = ρ_s/ρ_L → ships, swimmers, submarines, icebergs, fish and balloons
5.16 Key Terms
- Apparent weight — the weight a body seems to have while it is in a fluid, equal to its real weight minus the upthrust; measured in newton (N), and also in kgf or gf.
- Average density — the total mass of a body, counting both its material and any air enclosed inside it, divided by its total volume; measured in \(\\text{kg m}^{-3}\).
- Ballast — sand or stones placed at the bottom of an unloaded ship to lower its centre of gravity and make its equilibrium stable.
- Buoyancy — the property of a fluid by which it exerts an upward force on a body placed in it.
- Buoyant force — another name for the upthrust, the upward force a fluid exerts on a body immersed in it; measured in newton (N).
- Centre of buoyancy — the point at which the upthrust is taken to act, namely the centre of gravity of the displaced fluid.
- Density — the mass per unit volume of a substance; measured in \(\\text{kg m}^{-3}\) in SI units, or \(\\text{g cm}^{-3}\) in C.G.S. units.
- Displace — to push aside a volume of fluid equal to the volume of the submerged part of a body placed in it.
- Eureka can — a vessel with a side spout, used to collect and measure the liquid displaced by a body immersed in it.
- Fluid — any substance that can flow, that is, a liquid or a gas.
- Iceberg — a large floating mass of ice, which floats with nearly nine-tenths of its volume below the water surface.
- Lamina — a very thin sheet, whose upper and lower surfaces lie at almost the same depth, so that the liquid exerts a negligible upthrust on it.
- Plimsoll line — the white line painted on a ship’s side marking the safe limit for loading it in water of density \(10^{3}\\ \\text{kg m}^{-3}\).
- Pressure — the thrust acting on unit area of a surface; measured in pascal (Pa), where \(1\\ \\text{Pa} = 1\\ \\text{N m}^{-2}\).
- Relative density — the ratio of the density of a substance to the density of water at \(4^\\circ\\text{C}\); it is a pure ratio and has no unit.
- Swim bladder — the gas-filled organ of many fish, whose changing volume alters their average density and so lets them rise or sink in water.
- Upthrust — the upward force exerted on a body by the fluid in which it is submerged; measured in newton (N), and also in kgf or gf.
5.17 Points To Ponder
- The upthrust on a fully immersed body does not change as the body is taken deeper, even though the pressure on both its faces rises steadily. What does this tell you about which quantity in \(F\_B = V\\rho g\) is doing the real work, and why the pressure difference matters more than the pressure itself?
- A boat is floating in a small pond, carrying a heavy stone. The stone is thrown overboard and sinks to the bottom. Think through what happens to the weight of water displaced before and after, and decide whether the water level in the pond rises, falls, or stays the same.
- Archimedes’ principle is stated for a body immersed in a fluid, and we say that a hot-air balloon and a swimmer obey the same rule. Yet a stone lying flat on the muddy bottom of a tank, with no water underneath it, may not lift at all. Where exactly does the principle’s reasoning break down in that case?
- Relative density has no unit, while density has two common ones. Why does removing the unit make relative density more useful when comparing substances, and what information, if any, is lost in doing so?
- A ship built to sail in the North Atlantic will float at a different depth in tropical seas, and shipping rules take account of this. Using \(\\dfrac{v}{V} = \\dfrac{\\rho\_s}{\\rho\_L}\), work out which conditions would put a loaded ship deepest in the water, and consider why a single safe-loading mark cannot serve every ocean equally.
- A fish changes its swim bladder volume to rise and sink; a submarine changes the water in its ballast tanks. Both alter average density, but one adds mass while the other changes volume. Which of these two routes would be easier to reverse in an emergency, and why?
5.18 Objective Questions
- The upthrust on a body immersed in a liquid depends on: (1 mark)
- the mass of the body
- the material of the body
- the volume of the body submerged and the density of the liquid
- the depth of the body below the surface only
- A body of density \(\\rho\) is placed in a liquid of density \(\\rho\_L\). It will float with a part of it above the surface when: (1 mark)
- \(\\rho > \\rho\_L\)
- \(\\rho = \\rho\_L\)
- \(\\rho < \\rho\_L\)
- nothing can be said
- The apparent weight of a body floating freely on the surface of a liquid is: (1 mark)
- equal to its real weight
- zero
- greater than its real weight
- equal to the upthrust minus the weight
- The relative density of a substance is a ratio, and therefore: (1 mark)
- its unit is \(\\text{g cm}^{-3}\)
- its unit is \(\\text{kg m}^{-3}\)
- it has no unit
- its unit is \(\\text{m}^3\\text{ kg}^{-1}\)
- The density of water is maximum at: (1 mark)
- \(0^\\circ\\text{C}\)
- \(4^\\circ\\text{C}\)
- \(10^\\circ\\text{C}\)
- \(100^\\circ\\text{C}\)
- A ship sails from sea water into river water. It will: (1 mark)
- rise higher in the water
- sink deeper into the water
- float at exactly the same depth
- sink to the bottom
Choose from: a. Both A and R are true and R is the correct explanation of A b. Both A and R are true but R is not the correct explanation of A c. A is true but R is false d. A is false but R is true
Assertion (A): A bucket of water feels lighter while it is still inside the well than after it has come clear of the water. Reason (R): The water in the well exerts an upthrust on the bucket, which reduces its apparent weight. (1 mark)
Assertion (A): An iron ship floats on water although iron is much denser than water. Reason (R): The ship is hollow, so its average density is less than the density of water. (1 mark)
Assertion (A): A thin metal lamina placed flat in water sinks even though metal sheets are very light. Reason (R): The upper and lower surfaces of a lamina are at nearly the same depth, so the liquid gives it a negligible upthrust. (1 mark)
Match each item in Column A with the correct item in Column B. (1 mark)
| Column A | Column B |
|---|---|
| a. Upthrust | i. Has no unit |
| b. Relative density | ii. Marks the safe loading limit of a ship |
| c. Plimsoll line | iii. Measured in newton |
| d. Ballast | iv. Lowers the centre of gravity of a ship |
Fill in the blanks.
- The upthrust on a body acts vertically ………… at the centre of ………… . (1 mark)
- A body sinks in a liquid if its density is ………… than the density of the liquid. (1 mark)
- \(1\\ \\text{g cm}^{-3} = \\dots\\dots\\dots\\ \\text{kg m}^{-3}\). (1 mark)
State whether the following statements are true or false.
- The upthrust on a fully immersed body goes on increasing as it is taken deeper into the liquid. (1 mark)
- The mass of a body decreases when it is immersed in water. (1 mark)
Name the following.
- Name the force that acts vertically upwards on a body immersed in a fluid. (1 mark)
- Name the organ by which a fish alters its average density in water. (1 mark)
Give one word or phrase for each of the following.
- The point at which the upthrust on a body is taken to act. (1 mark)
- The weight a body appears to have when it is placed in a fluid. (1 mark)
State the SI unit of (i) upthrust, (ii) density, and (iii) relative density. (1 mark)
Cork, wood (pine), petrol, mercury. Pick the odd one out and give a reason for your choice. (1 mark)
5.19 Check Your Concepts
Define upthrust and state its SI unit. (2 marks)
State Archimedes’ principle. (2 marks)
State the two factors on which the upthrust on a body immersed in a fluid depends. (2 marks)
Give a reason why a body weighed on a sensitive spring balance weighs slightly less in air than in a vacuum. (2 marks)
Differentiate between density and relative density, stating the basis of difference in each case. (3 marks)
Explain why an iron nail sinks in water while a ship built of iron floats on it. (3 marks)
State the principle of floatation, and use it to explain why a loaded ship floats lower in the water than the same ship when unloaded. (3 marks)
Why does a swimmer find it easier to float in sea water than in river water? Support your answer using the relation \(\\dfrac{v}{V} = \\dfrac{\\rho\_s}{\\rho\_L}\). (3 marks)
Explain, using pressure, why a body immersed in a liquid experiences an upward force. Why do the thrusts on its vertical sides not contribute to this force? (4 marks)
Describe an experiment, using a eureka can and a spring balance, to verify Archimedes’ principle. State the observations you would take and the conclusion you would draw. (5 marks)
What would happen to a submarine if water were let into its ballast tanks while it floated on the surface? Explain the change in terms of average density and upthrust, and state how the change could be reversed. (4 marks)
A block of wood floats in water and the same block floats in glycerine. In which liquid is a greater volume of the block submerged? Give a reason for your answer. (3 marks)
5.20 Diagrams, Graphs & Constructions
Fig. 5.1-2 shows a body held completely immersed in a liquid. Redraw the diagram, mark the two forces acting on the body with their directions, and state the condition under which the body would sink on being released. (3 marks)
Fig. 5.12-1 shows three cases of a block placed in a liquid. Identify which panel represents a floating body of density equal to that of the liquid, and give a reason for your choice. (3 marks)
Draw a labelled diagram of a floating body showing its centre of gravity \(G\) and its centre of buoyancy \(B\), and mark the weight and the upthrust with their points of application. (3 marks)
Fig. 5.8-1 shows how the density of water varies with temperature. Read the graph and state (i) the temperature at which the density is greatest, (ii) how the density changes between \(0^\\circ\\text{C}\) and \(4^\\circ\\text{C}\), and (iii) what the shape of the curve above \(4^\\circ\\text{C}\) tells you about the volume of a fixed mass of water. (4 marks)
Draw a labelled diagram of the apparatus you would set up, as in Fig. 5.10-1, to find the relative density of a solid denser than water and insoluble in it. State one precaution to be taken while immersing the solid. (4 marks)
Sketch, without exact values, the shape of the graph you would expect if the upthrust on a block were plotted against the volume of the block submerged in a given liquid, up to the point of full immersion. Label both axes with their units, and state what the gradient of the line represents. (4 marks)
5.21 Practice With Numericals
A body of volume \(60\\ \\text{cm}^3\) is completely immersed in water of density \(1000\\ \\text{kg m}^{-3}\). Calculate the upthrust on the body. Take \(g = 10\\ \\text{m s}^{-2}\). (2 marks)
A solid weighs \(96\\ \\text{gf}\) in air and \(84\\ \\text{gf}\) when completely immersed in water of density \(1\\ \\text{g cm}^{-3}\). Find (i) the loss in weight of the solid, (ii) the upthrust on it, and (iii) the volume of the solid. (3 marks)
A body of mass \(250\\ \\text{g}\) has a volume of \(200\\ \\text{cm}^3\). Calculate its density and state, with a reason, whether it will float or sink in water of density \(1\\ \\text{g cm}^{-3}\). (3 marks)
The density of mercury is \(13600\\ \\text{kg m}^{-3}\). Find (i) its relative density and (ii) its density in \(\\text{g cm}^{-3}\). Take the density of water as \(1000\\ \\text{kg m}^{-3}\). (2 marks)
A solid weighs \(150\\ \\text{gf}\) in air and \(125\\ \\text{gf}\) when completely immersed in water. Calculate the relative density of the solid. (3 marks)
A metal piece weighs \(40\\ \\text{gf}\) in air, \(33\\ \\text{gf}\) when completely immersed in a liquid and \(30\\ \\text{gf}\) when completely immersed in water. Calculate the relative density of the liquid. (3 marks)
A block of volume \(800\\ \\text{cm}^3\) and density \(0.75\\ \\text{g cm}^{-3}\) floats freely in water of density \(1\\ \\text{g cm}^{-3}\). Find the volume of the block below the water surface and the volume above it. (3 marks)
A piece of ice of density \(0.9\\ \\text{g cm}^{-3}\) floats in a liquid of density \(1.2\\ \\text{g cm}^{-3}\). Find the percentage of its volume that remains above the surface of the liquid. (3 marks)
A block of mass \(3.2\\ \\text{kg}\) and volume \(0.004\\ \\text{m}^3\) floats freely in water of density \(1000\\ \\text{kg m}^{-3}\). Taking \(g = 10\\ \\text{m s}^{-2}\), find (i) the weight of water displaced and (ii) the volume of the block below the water surface. (4 marks)
A solid of relative density \(8.5\) weighs \(1.7\\ \\text{N}\) in air. Calculate its weight when it is completely immersed in water. (4 marks)
A balloon of volume \(200\\ \\text{m}^3\) is filled with a gas of density \(0.2\\ \\text{kg m}^{-3}\). The mass of the empty balloon is \(60\\ \\text{kg}\) and the density of the surrounding air is \(1.3\\ \\text{kg m}^{-3}\). Calculate the maximum load, in kgf, that the balloon can lift. (4 marks)