A Stone Loses Weight the Moment It Touches Water, and You Can Say Exactly How Much
Learn what causes upthrust and what it depends on, calculate apparent weight and apparent loss in weight, state Archimedes' principle and verify it, and use upthrust equal to V rho g in numericals.
Why does a stone feel lighter under water?
Hang a stone from a spring balance and it reads N. Lower the stone into water, still hanging, and the reading drops to N.
The stone did not change. Nothing was removed from it, and on dry land it would still read N.
What has happened is that the water is pushing upward on the stone with a force of N, so the spring only has to support the remaining N. That upward push is called upthrust, and the N is the apparent loss in weight.
Where does the push come from? From the fact established two chapters ago that liquid pressure increases with depth. The bottom face of the stone is deeper than the top face, so the water presses up on the bottom harder than it presses down on the top. The difference is a net upward force.
So upthrust is not a new law of nature bolted on for floating objects. **It is applied to two faces of the same body**, and the next section derives it that way in three lines.
This page covers the first part of the ICSE Class 9 Physics chapter on upthrust and floatation — what causes upthrust, apparent weight, Archimedes' principle, and the formula .
The stone did not change. Nothing was removed from it, and on dry land it would still read N.
What has happened is that the water is pushing upward on the stone with a force of N, so the spring only has to support the remaining N. That upward push is called upthrust, and the N is the apparent loss in weight.
Where does the push come from? From the fact established two chapters ago that liquid pressure increases with depth. The bottom face of the stone is deeper than the top face, so the water presses up on the bottom harder than it presses down on the top. The difference is a net upward force.
So upthrust is not a new law of nature bolted on for floating objects. **It is applied to two faces of the same body**, and the next section derives it that way in three lines.
This page covers the first part of the ICSE Class 9 Physics chapter on upthrust and floatation — what causes upthrust, apparent weight, Archimedes' principle, and the formula .
What causes upthrust, and what does it depend on?
Upthrust is caused by the pressure on the lower face of an immersed body exceeding the pressure on its upper face, and the difference multiplied by the area is a net upward force.
Buoyancy is the tendency of a fluid to exert that upward force; upthrust, or the buoyant force, is the force itself. Its SI unit is the newton.
The derivation. Take a cube of side immersed in a liquid of density , with its top face at depth and its bottom face at . Each face has area .
- Downward thrust on the top face:
- Upward thrust on the bottom face:
The side faces are pushed inward equally from opposite directions, so they cancel. The net upward force is
and is the cube's volume, so
**And is the mass of the liquid the cube pushed aside**, so is the weight of the liquid displaced — which is Archimedes' principle, arrived at by arithmetic rather than asserted.
Worked example. A cube of side m, so , fully immersed in water:
Upthrust depends on two things only.
- The volume of the body immersed — more volume displaces more liquid
- The density of the liquid — a denser liquid weighs more per unit volume
**And on , which is usually the same throughout a problem.
Upthrust does NOT depend on:
- The depth**, for a body fully immersed in a uniform liquid. The derivation used , and the individual depths cancelled — so lowering the cube further changes nothing
- The mass or the material of the body. An iron cube and a wooden cube of the same size feel the same upthrust in water
- The body's own density
Worked comparison. An iron cube and a wooden cube, both of side m, fully immersed in water. Both experience N of upthrust. The iron one sinks and the wooden one floats, and the upthrust on each while fully immersed is identical — what differs is their weights, which is the comparison the next part of this chapter makes into a rule.
Only the immersed volume counts. A body floating with half its volume under water displaces half as much liquid and receives half the upthrust. So "volume of the body" is wrong and "volume immersed" is right, and for a fully submerged body they happen to coincide — which is why the distinction only shows up in floating problems.
Buoyancy is the tendency of a fluid to exert that upward force; upthrust, or the buoyant force, is the force itself. Its SI unit is the newton.
The derivation. Take a cube of side immersed in a liquid of density , with its top face at depth and its bottom face at . Each face has area .
- Downward thrust on the top face:
- Upward thrust on the bottom face:
The side faces are pushed inward equally from opposite directions, so they cancel. The net upward force is
and is the cube's volume, so
**And is the mass of the liquid the cube pushed aside**, so is the weight of the liquid displaced — which is Archimedes' principle, arrived at by arithmetic rather than asserted.
Worked example. A cube of side m, so , fully immersed in water:
Upthrust depends on two things only.
- The volume of the body immersed — more volume displaces more liquid
- The density of the liquid — a denser liquid weighs more per unit volume
**And on , which is usually the same throughout a problem.
Upthrust does NOT depend on:
- The depth**, for a body fully immersed in a uniform liquid. The derivation used , and the individual depths cancelled — so lowering the cube further changes nothing
- The mass or the material of the body. An iron cube and a wooden cube of the same size feel the same upthrust in water
- The body's own density
Worked comparison. An iron cube and a wooden cube, both of side m, fully immersed in water. Both experience N of upthrust. The iron one sinks and the wooden one floats, and the upthrust on each while fully immersed is identical — what differs is their weights, which is the comparison the next part of this chapter makes into a rule.
Only the immersed volume counts. A body floating with half its volume under water displaces half as much liquid and receives half the upthrust. So "volume of the body" is wrong and "volume immersed" is right, and for a fully submerged body they happen to coincide — which is why the distinction only shows up in floating problems.
Formula
How do you calculate apparent weight and apparent loss in weight?
Apparent weight is the true weight less the upthrust, and the apparent loss in weight IS the upthrust.
Worked example 1 — reading the upthrust off a balance. A body weighs N in air and N in water:
and since , its volume is
So weighing a body twice measures its volume, whatever its shape — which is what makes this a useful laboratory method for an irregular solid.
Worked example 2 — from the volume instead. A metal block of volume weighs N in air. Convert the volume: .
Worked example 3 — half immersed. The same block with only half its volume under water:
Half the immersed volume, half the upthrust.
Worked example 4 — a body that will sink. A solid of mass g and volume is fully immersed in water.
The apparent weight is still positive, so the body sinks. Its density is , twice that of water — which is why.
Worked example 5 — in a different liquid. The block of worked example 2 immersed in a liquid of density :
Less upthrust in a lighter liquid, so the body feels heavier than it did in water.
Worked example 6 — finding the liquid's density. A body weighs N in air, N in water and N in oil. From the water reading, N; from the oil reading, N. Since with the same and ,
The volume never had to be found — it cancelled in the ratio, and that shortcut is the basis of the relative-density method in the next part of this chapter.
**Convert to by dividing by , not by .** Since m, cubing gives — the raise-the-factor rule from the measurements chapter. **Using makes every upthrust ten thousand times too large**, and it is the most frequent numerical error in this chapter.
Worked example 1 — reading the upthrust off a balance. A body weighs N in air and N in water:
and since , its volume is
So weighing a body twice measures its volume, whatever its shape — which is what makes this a useful laboratory method for an irregular solid.
Worked example 2 — from the volume instead. A metal block of volume weighs N in air. Convert the volume: .
Worked example 3 — half immersed. The same block with only half its volume under water:
Half the immersed volume, half the upthrust.
Worked example 4 — a body that will sink. A solid of mass g and volume is fully immersed in water.
The apparent weight is still positive, so the body sinks. Its density is , twice that of water — which is why.
Worked example 5 — in a different liquid. The block of worked example 2 immersed in a liquid of density :
Less upthrust in a lighter liquid, so the body feels heavier than it did in water.
Worked example 6 — finding the liquid's density. A body weighs N in air, N in water and N in oil. From the water reading, N; from the oil reading, N. Since with the same and ,
The volume never had to be found — it cancelled in the ratio, and that shortcut is the basis of the relative-density method in the next part of this chapter.
**Convert to by dividing by , not by .** Since m, cubing gives — the raise-the-factor rule from the measurements chapter. **Using makes every upthrust ten thousand times too large**, and it is the most frequent numerical error in this chapter.
What does Archimedes' principle say, and how do you verify it?
When a body is immersed wholly or partially in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.
The derivation in the second section already produced this: , where is the mass of liquid pushed aside. The principle is the derivation's conclusion stated in words.
The experiment to verify it.
- Weigh the solid in air with a spring balance. Call the reading
- Fill an overflow can (a spouted vessel) with water right up to the level of the spout, and let it stop dripping
- Place a dry, pre-weighed beaker under the spout
- Lower the solid, still hanging from the balance, until it is fully immersed, and note the new reading . Collect all the overflow in the beaker
- Weigh the beaker with the collected water and subtract the empty beaker's weight to get the weight of the displaced water,
The result. Within experimental error,
The apparent loss in weight equals the weight of the water displaced.
Worked example — the numbers from such an experiment. A solid weighs N in air and N when fully immersed, and the displaced water is found to weigh N.
The principle is confirmed. And the solid's volume follows:
Three precautions that decide whether the experiment works.
- The can must be filled exactly to the spout and allowed to stop dripping first, or some overflow is lost or added
- The solid must be fully immersed without touching the sides or the bottom of the can, since a touch would let the can support part of its weight
- No air bubbles should cling to the solid, as a bubble adds volume and inflates the upthrust
The principle covers gases too. Archimedes' principle says fluid, not liquid, so a body in air also receives an upthrust equal to the weight of the air it displaces. For a stone that upthrust is negligible; for a hydrogen balloon, which displaces a large volume of air, it is what lifts it. So a balloon rising and a cork rising in water are the same physics, and the word fluid in the statement is doing real work.
The principle applies to a partly immersed body as well, with being the immersed volume. That case is what floating is, and it is the subject of the next part of this chapter — where a floating body turns out to displace exactly its own weight of liquid.
The derivation in the second section already produced this: , where is the mass of liquid pushed aside. The principle is the derivation's conclusion stated in words.
The experiment to verify it.
- Weigh the solid in air with a spring balance. Call the reading
- Fill an overflow can (a spouted vessel) with water right up to the level of the spout, and let it stop dripping
- Place a dry, pre-weighed beaker under the spout
- Lower the solid, still hanging from the balance, until it is fully immersed, and note the new reading . Collect all the overflow in the beaker
- Weigh the beaker with the collected water and subtract the empty beaker's weight to get the weight of the displaced water,
The result. Within experimental error,
The apparent loss in weight equals the weight of the water displaced.
Worked example — the numbers from such an experiment. A solid weighs N in air and N when fully immersed, and the displaced water is found to weigh N.
The principle is confirmed. And the solid's volume follows:
Three precautions that decide whether the experiment works.
- The can must be filled exactly to the spout and allowed to stop dripping first, or some overflow is lost or added
- The solid must be fully immersed without touching the sides or the bottom of the can, since a touch would let the can support part of its weight
- No air bubbles should cling to the solid, as a bubble adds volume and inflates the upthrust
The principle covers gases too. Archimedes' principle says fluid, not liquid, so a body in air also receives an upthrust equal to the weight of the air it displaces. For a stone that upthrust is negligible; for a hydrogen balloon, which displaces a large volume of air, it is what lifts it. So a balloon rising and a cork rising in water are the same physics, and the word fluid in the statement is doing real work.
The principle applies to a partly immersed body as well, with being the immersed volume. That case is what floating is, and it is the subject of the next part of this chapter — where a floating body turns out to displace exactly its own weight of liquid.
How do you use upthrust equal to V rho g in numericals?
**Identify which of , and is missing, then rearrange.** All three forms get used.
Worked example 1 — finding the upthrust. A body of volume is fully immersed in a liquid of density . Converting, :
Worked example 2 — finding the volume. A body experiences an upthrust of N when fully immersed in water:
Worked example 3 — finding the liquid's density. A body of volume receives N of upthrust when fully immersed:
Worked example 4 — a floating body. A block of volume floats in water with submerged. The upthrust uses only the immersed volume:
Since the block floats at rest, the upthrust must balance its weight exactly:
and its density is
Notice what came out of that. The block's density, , divided by the water's density, , gives — and is also . The fraction submerged equals the ratio of the densities, a result the next part of this chapter derives properly.
Worked example 5 — an iron block in mercury. A block of volume and mass g is placed in mercury of density . Fully immersed, the upthrust would be
against a weight of N. The upthrust exceeds the weight, so the block cannot stay submerged — it rises and floats, which is why iron floats on mercury.
Worked example 6 — two liquids compared. The same body fully immersed:
- in water: N
- in mercury: N
**The ratio is **, exactly the ratio of the densities — the upthrust is proportional to with everything else fixed.
Check whether the body is fully or partly immersed before substituting. For a sinking body is the whole volume; for a floating one it is only the part below the surface. Using the full volume for a floating body is the error that produces an upthrust larger than the weight — which is physically impossible for a body sitting still, and is therefore its own warning sign.
Worked example 1 — finding the upthrust. A body of volume is fully immersed in a liquid of density . Converting, :
Worked example 2 — finding the volume. A body experiences an upthrust of N when fully immersed in water:
Worked example 3 — finding the liquid's density. A body of volume receives N of upthrust when fully immersed:
Worked example 4 — a floating body. A block of volume floats in water with submerged. The upthrust uses only the immersed volume:
Since the block floats at rest, the upthrust must balance its weight exactly:
and its density is
Notice what came out of that. The block's density, , divided by the water's density, , gives — and is also . The fraction submerged equals the ratio of the densities, a result the next part of this chapter derives properly.
Worked example 5 — an iron block in mercury. A block of volume and mass g is placed in mercury of density . Fully immersed, the upthrust would be
against a weight of N. The upthrust exceeds the weight, so the block cannot stay submerged — it rises and floats, which is why iron floats on mercury.
Worked example 6 — two liquids compared. The same body fully immersed:
- in water: N
- in mercury: N
**The ratio is **, exactly the ratio of the densities — the upthrust is proportional to with everything else fixed.
Check whether the body is fully or partly immersed before substituting. For a sinking body is the whole volume; for a floating one it is only the part below the surface. Using the full volume for a floating body is the error that produces an upthrust larger than the weight — which is physically impossible for a body sitting still, and is therefore its own warning sign.
Exam tip
Exam tip: convert cubic centimetres properly and use the immersed volume
****, not . Divide by a million: .
**Use the IMMERSED volume in — the whole volume only for a fully submerged body.
Apparent weight , and the apparent LOSS in weight IS the upthrust. Write both lines.
Weighing twice gives the volume**: , whatever the shape.
For two liquids, take the ratio of the upthrusts — the volume cancels, so no volume is needed: gives .
Upthrust depends only on the immersed volume and the liquid's density — not on depth, mass, material or the body's own density. Say all of that when asked.
For the derivation, show the two thrusts and the subtraction: becomes . The cancelling of the side faces is worth one line.
State Archimedes' principle with the word FLUID, not liquid — it covers gases, which is why a balloon rises.
List the experiment's precautions: filled to the spout, fully immersed without touching, and no air bubbles.
For a floating body, weight equals upthrust — that equality is what "floating at rest" means.
And if an upthrust comes out greater than the weight for a body at rest, a volume was wrong — check whether the body was floating.
**Use the IMMERSED volume in — the whole volume only for a fully submerged body.
Apparent weight , and the apparent LOSS in weight IS the upthrust. Write both lines.
Weighing twice gives the volume**: , whatever the shape.
For two liquids, take the ratio of the upthrusts — the volume cancels, so no volume is needed: gives .
Upthrust depends only on the immersed volume and the liquid's density — not on depth, mass, material or the body's own density. Say all of that when asked.
For the derivation, show the two thrusts and the subtraction: becomes . The cancelling of the side faces is worth one line.
State Archimedes' principle with the word FLUID, not liquid — it covers gases, which is why a balloon rises.
List the experiment's precautions: filled to the spout, fully immersed without touching, and no air bubbles.
For a floating body, weight equals upthrust — that equality is what "floating at rest" means.
And if an upthrust comes out greater than the weight for a body at rest, a volume was wrong — check whether the body was floating.
Did you know
Why an iron block floats on mercury and sinks in water
Put a solid iron block in a bowl of water and it goes straight to the bottom. Put the same block in a dish of mercury and it sits on the surface like a cork.
Nothing about the iron changed. What changed is what it was asked to displace.
Take a block of volume and mass g, so a weight of N.
In water, fully immersed, it displaces of water, which has a mass of g:
The upthrust is N against a weight of N — nowhere near enough, so the block sinks.
In mercury, the same displaced has a mass of g:
Now the upthrust available exceeds the weight, and the block cannot stay under. It rises until only part of it is submerged, settling where the displaced mercury weighs exactly N — which is g of mercury, occupying about . So a little over half the block sits below the surface.
The whole comparison is one number: **iron is about times denser than water and about times as dense as mercury.** Sinking and floating are not properties of a material on their own; they are comparisons between two materials.
Which is why the question does iron float has no answer until the liquid is named. Iron sinks in water, floats on mercury, and would sink in almost anything you could pour — and a hollow iron ship floats on water, because hollowing it out changes the average density without changing the iron at all.
Nothing about the iron changed. What changed is what it was asked to displace.
Take a block of volume and mass g, so a weight of N.
In water, fully immersed, it displaces of water, which has a mass of g:
The upthrust is N against a weight of N — nowhere near enough, so the block sinks.
In mercury, the same displaced has a mass of g:
Now the upthrust available exceeds the weight, and the block cannot stay under. It rises until only part of it is submerged, settling where the displaced mercury weighs exactly N — which is g of mercury, occupying about . So a little over half the block sits below the surface.
The whole comparison is one number: **iron is about times denser than water and about times as dense as mercury.** Sinking and floating are not properties of a material on their own; they are comparisons between two materials.
Which is why the question does iron float has no answer until the liquid is named. Iron sinks in water, floats on mercury, and would sink in almost anything you could pour — and a hollow iron ship floats on water, because hollowing it out changes the average density without changing the iron at all.
Exam relevance
How is upthrust tested in JEE Main and NEET?
Because Archimedes' principle is the basis of every floatation problem, and the pressure-difference derivation is the standard proof asked for.
This is the foundation for Class 11 Physics Mechanical Properties of Fluids, examined in JEE Main and NEET. The derivation performed on this page — pressure on the lower face minus pressure on the upper face, times the area — is exactly the proof given there, generalised from a cube to any shape. **Recognising that upthrust is a consequence of rather than a separate law is what that chapter builds on.
Apparent weight problems extend to accelerating frames.** In a lift accelerating upward, the effective value of becomes , and both the weight and the upthrust scale by the same factor — so the apparent weight of a submerged body changes but the fraction submerged for a floating body does not. That is a favourite JEE Main assertion-reason item, and it works because and contain the same .
The two-liquid ratio method used in worked example 6 of the third section becomes the standard laboratory determination of relative density, which is the next part of this chapter, and the same cancellation trick appears in Class 11 numericals for finding an unknown density.
Where the gas case matters. The point that Archimedes' principle says fluid is used in Class 11 for balloons and in Class 11 Chemistry when weighing a gas — a body weighed in air reads slightly less than its true weight because of the air's upthrust, and correcting for that is a buoyancy correction. NEET uses the same principle for the flotation of organisms and for the swim bladder.
Surface tension and viscosity are added in Class 11 as the other forces on an immersed body, and a terminal-velocity problem balances three forces — weight, upthrust and viscous drag. **The upthrust term in that balance is from this page, unchanged, and terminal-velocity numericals are recurring in both exams.
What the questions look like. For board work, expect define buoyancy and upthrust, state the factors it depends on and those it does not, derive upthrust from the pressure difference, calculate apparent weight and apparent loss, state Archimedes' principle and describe its verification with precautions, and numericals using in all three rearrangements. For JEE Main and NEET, expect apparent weight in accelerating frames, terminal velocity, buoyancy corrections and floatation problems.
How board and competitive emphasis differ. A board paper rewards the derivation with the side faces shown to cancel** and the experiment's precautions listed. A competitive paper assumes and tests whether it is used with the right volume in a force balance — often alongside two other forces.
The single trap that costs the most marks. Converting to by dividing by instead of by a million. A volume of is , and using gives an upthrust ten thousand times too large. **The defence is to write explicitly** so the cube lands on the factor and not only on the unit — the same discipline the measurements chapter demanded for density.
This is the foundation for Class 11 Physics Mechanical Properties of Fluids, examined in JEE Main and NEET. The derivation performed on this page — pressure on the lower face minus pressure on the upper face, times the area — is exactly the proof given there, generalised from a cube to any shape. **Recognising that upthrust is a consequence of rather than a separate law is what that chapter builds on.
Apparent weight problems extend to accelerating frames.** In a lift accelerating upward, the effective value of becomes , and both the weight and the upthrust scale by the same factor — so the apparent weight of a submerged body changes but the fraction submerged for a floating body does not. That is a favourite JEE Main assertion-reason item, and it works because and contain the same .
The two-liquid ratio method used in worked example 6 of the third section becomes the standard laboratory determination of relative density, which is the next part of this chapter, and the same cancellation trick appears in Class 11 numericals for finding an unknown density.
Where the gas case matters. The point that Archimedes' principle says fluid is used in Class 11 for balloons and in Class 11 Chemistry when weighing a gas — a body weighed in air reads slightly less than its true weight because of the air's upthrust, and correcting for that is a buoyancy correction. NEET uses the same principle for the flotation of organisms and for the swim bladder.
Surface tension and viscosity are added in Class 11 as the other forces on an immersed body, and a terminal-velocity problem balances three forces — weight, upthrust and viscous drag. **The upthrust term in that balance is from this page, unchanged, and terminal-velocity numericals are recurring in both exams.
What the questions look like. For board work, expect define buoyancy and upthrust, state the factors it depends on and those it does not, derive upthrust from the pressure difference, calculate apparent weight and apparent loss, state Archimedes' principle and describe its verification with precautions, and numericals using in all three rearrangements. For JEE Main and NEET, expect apparent weight in accelerating frames, terminal velocity, buoyancy corrections and floatation problems.
How board and competitive emphasis differ. A board paper rewards the derivation with the side faces shown to cancel** and the experiment's precautions listed. A competitive paper assumes and tests whether it is used with the right volume in a force balance — often alongside two other forces.
The single trap that costs the most marks. Converting to by dividing by instead of by a million. A volume of is , and using gives an upthrust ten thousand times too large. **The defence is to write explicitly** so the cube lands on the factor and not only on the unit — the same discipline the measurements chapter demanded for density.
Key takeaways
Upthrust, apparent weight and Archimedes' principle: quick revision
- Buoyancy is a fluid's tendency to push an immersed body up; upthrust is that force, in newtons.
- Cause: liquid pressure increases with depth, so the bottom face is pushed up harder than the top face is pushed down; the side faces cancel.
- Derivation for a cube of side : , and is the mass of liquid displaced.
- A cube of side m () in water gets N.
- Upthrust depends on the immersed volume and the liquid's density (and ).
- It does NOT depend on the depth (for full immersion), the mass, the material, or the body's own density — an iron and a wooden cube of equal size get equal upthrust.
- **Apparent weight , and the apparent loss in weight IS the upthrust**.
- N in air and N in water gives N and — weighing twice measures the volume.
- A block weighing N: N fully immersed, so apparent weight N; half immersed, N and N.
- A g, solid: N, N, apparent weight N. Its density is , so it sinks.
- In a liquid of the block gets N, so it feels heavier than in water.
- Two-liquid ratio: N in air, N in water, N in oil gives , so the oil is — the volume cancels.
- **** — divide by a million, not by .
- Archimedes' principle: a body immersed wholly or partially in a fluid experiences an upthrust equal to the weight of the fluid displaced.
- Verification: weigh in air (), weigh fully immersed (), collect and weigh the overflow (), and find . With N, N and N it checks exactly, giving .
- Precautions: fill to the spout and let it stop dripping, immerse fully without touching, and remove air bubbles.
- It says fluid, not liquid — a balloon rises because it displaces a large weight of air.
- Numericals: in a liquid of gives N; N in water means ; with N means .
- A block floating with under water has N, so g and — the fraction submerged equals the density ratio.
- A , g iron block gets N in water (sinks) and N in mercury (floats) — **the ratio is the density ratio.
- For a floating body, weight equals upthrust**; an upthrust bigger than the weight means the wrong volume was used.
Weigh a stone on a kitchen scale, then weigh it again hanging in a jug of water, and work out its volume from the difference — no measuring cylinder needed.
- Cause: liquid pressure increases with depth, so the bottom face is pushed up harder than the top face is pushed down; the side faces cancel.
- Derivation for a cube of side : , and is the mass of liquid displaced.
- A cube of side m () in water gets N.
- Upthrust depends on the immersed volume and the liquid's density (and ).
- It does NOT depend on the depth (for full immersion), the mass, the material, or the body's own density — an iron and a wooden cube of equal size get equal upthrust.
- **Apparent weight , and the apparent loss in weight IS the upthrust**.
- N in air and N in water gives N and — weighing twice measures the volume.
- A block weighing N: N fully immersed, so apparent weight N; half immersed, N and N.
- A g, solid: N, N, apparent weight N. Its density is , so it sinks.
- In a liquid of the block gets N, so it feels heavier than in water.
- Two-liquid ratio: N in air, N in water, N in oil gives , so the oil is — the volume cancels.
- **** — divide by a million, not by .
- Archimedes' principle: a body immersed wholly or partially in a fluid experiences an upthrust equal to the weight of the fluid displaced.
- Verification: weigh in air (), weigh fully immersed (), collect and weigh the overflow (), and find . With N, N and N it checks exactly, giving .
- Precautions: fill to the spout and let it stop dripping, immerse fully without touching, and remove air bubbles.
- It says fluid, not liquid — a balloon rises because it displaces a large weight of air.
- Numericals: in a liquid of gives N; N in water means ; with N means .
- A block floating with under water has N, so g and — the fraction submerged equals the density ratio.
- A , g iron block gets N in water (sinks) and N in mercury (floats) — **the ratio is the density ratio.
- For a floating body, weight equals upthrust**; an upthrust bigger than the weight means the wrong volume was used.
Weigh a stone on a kitchen scale, then weigh it again hanging in a jug of water, and work out its volume from the difference — no measuring cylinder needed.