# Upthrust Archimedes Principle Floatation Selina Class-9

## ICSE Class-9 Physics Revised Concise Selina Chapter-5

Upthrust Archimedes Principle Floatation Selina Concise Class-9 Physics Chapter-5 .Step By Step Revised Concise Selina Physics Solutions of Chapter-5 **Upthrust in Fluids, Archimedes’ Principle and Floatation** with Exe-5(A) , Numericals-5(A) , MCQ 5(A), Exe-5(B) , Numericals-5(B), MCQ -5(B), Exe-5(C) Numericals -5(C) and MCQ-5(C) for Class-9. Visit official Website **CISCE ** for detail information about ICSE Board Class-9..

## Upthrust Archimedes Principle Floatation Selina Concise Class-9 Physics solutions

The chapter **Upthrust in Fluids**, is very important in physics **Archimedes’ Principle** is based on **law of flotation.** In the exam of council this chapter question has been asked every year before solving the chapter **Upthrust in Fluids**, you should Know about the **basic of Upthrust, Archimedes’ principle, Law of Floatation **and about density of matter. The Application of principal of floatation is also use full to under stand this chapter completely.

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Note :- Before Viewing Selina Concise Physics Solutions of Chapter-5 **Upthrust in Fluids, Archimedes’ Principle and Floatation** for ICSE Class-9 Physics . Read the whole chapter carefully and Solved all example of Chapter-5 **Upthrust in Fluids, Archimedes’ Principle and Floatation** for Class-9 Physics**.**

**Exercise Ex 5(A) Upthrust in Fluids, Archimedes’ Principle and Floatation**

**Question 1**

What do you understand by the term upthrust of a fluid? Describe an experiment to show its existence.

**Answer 1**

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

##### Upthrust can be demonstrated by the following experiment:

Take an empty can and close its mouth with an airtight stopper. Put it in a tub filled with water. It floats with a large part of it above the surface of water and only a small part of it below the surface of water. Push the can into the water. You can feel an upward force and you find it difficult to push the can further into water.

It is noticed that as the can is pushed more and more into the water, more and more force is needed to push the can further into water, until it is completely immersed. When the can is fully inside the water, a definite force is still needed to keep it at rest in that position. Again, if the can is released in this position, it is noticed that the can bounces back to the surface and starts floating again.

**Question 2**

In what direction and at what point does the buoyant force on a body due to a liquid act?

**Answer 2**

Buoyant force on a body due to a liquid acts upwards at the centre of buoyancy.

**Question 3**

What is meant by the term buoyancy?

**Answer 3**

The property of a liquid to exert an upward force on a body immersed in it is called buoyancy.

**Question 4**

Define upthrust and state its S.I. unit.

**Answer 4**

The upward force exerted on a body by the fluid in which it is submerged is called the upthrust. Its S.I. unit is ‘newton’.

**Question 5**

What is the cause of upthrust? At which point can it be considered to act?

**Answer 5**

A liquid contained in a vessel exerts pressure at all points and in all directions. The pressure at a point in a liquid is the same in all directions – upwards, downwards and sideways. It increases with the depth inside the liquid.

When a body is immersed in a liquid, the thrusts acting on the side walls of the body are neutralized as they are equal in magnitude and opposite in direction. However, the magnitudes of pressure on the upper and lower faces are not equal. The difference in pressure on the upper and lower faces cause a net upward force (= pressure x area) or upthrust on the body.

It acts at the center of buoyancy.

**Question 6**

Why is a force needed to keep a block of wood inside water?

**Answer 6**

Upthrust due to water on block when fully submerged is more than its weight. Density of water is more than the density of cork; hence, upthrust due to water on the block of cork when fully submerged in water is more than its weight.

**Question 7**

A piece of wood if left under water comes to the surface. Explain the reason.

**Answer 7**

A piece of wood if left under water comes to the surface of water because the upthrust on body due to its submerged part is equal to its own weight.

**Question 8**

Describe an experiment to show that a body immersed in a liquid appears lighter than it really is.

**Answer 8**

Experiment to show that a body immersed in a liquid appears lighter:

Take a solid body and suspend it by a thin thread from the hook of a spring balance as shown in the above figure (a). Note its weight. Above figure (a) shows the weight as 0.67 N.

Then, take a can filled with water. Immerse the solid gently into the water while hanging from the hook of the spring balance as shown in figure (b). Note its weight. Above figure (b) shows the weight as 0.40 N.

The reading in this case (b) shall be less than the reading in the case (a), which proves that a body immersed in a liquid appears to be lighter.

**Question 9**

Will a body weigh more in air or vacuum when weighed with a spring balance? Give a reason for your answer.

**Answer 9**

A body shall weigh more in vacuum because in vacuum, i.e. in absence of air, no upthrust will act on the body.

**Question 10**

A metal solid cylinder tied to a thread is hanging from the hook of a spring balance. The cylinder is gradually immersed into the water contained in a jar. What changes do you expect in the readings of the spring balance? Explain your answer.

**Answer 10**

The readings in the spring balance decreases.

As the cylinder is immersed in the jar of water, an upward force acts on it, which is in opposition to the weight component of the cylinder. Hence the cylinder appears to be lighter.

**Question 11**

A body dipped into a liquid experiences an upthrust. State two factors on which upthrust on the body depends.

**Answer 11**

Upthrust on a body depends on the following factors:

(i) Volume of the body submerged in the liquid or fluid.

(ii) Density of liquid or fluid in which the body is submerged.

**Question 12**

How is the upthrust related to the volume of the body submerged in a liquid?

**Answer 12**

Larger the volume of body submerged in liquid, greater is the upthrust acting on it.

**Question 13**

A bunch of feathers and a stone of the same mass are released simultaneously in air. Which will fall faster and why? How will your observation be different if they are released simultaneously in vacuum?

**Answer 13**

When a bunch of feathers and a stone of the same mass are released simultaneously in air, the feathers will fall after the stone falls due to air friction. In vacuum, as there is no air friction, the acceleration due to gravity of both bodies will be the same, and therefore, the feathers and the stone will fall at the same time.

**Question 14**

A body experiences an upthrust F_{1} in river water and F_{2} in sea water when dipped up to the same level. Which is more, F_{1} or F_{2}? Give reason.

**Answer 14**

F_{2 }F_{1}; Sea water is denser than river water; therefore, the upthrust due to sea water will be greater than that due to river water at the same level. This shall make the body to appear lighter in the sea water.

**Question 15**

A small block of wood is completely immersed in (i) water, (ii) glycerine and then released. In each case, What do you observe? Explain the difference in your observation in the two cases.

**Answer 15**

Observation: Volume of a block of wood immersed in glycerin is smaller as compared to the volume of block immersed in water.

Explanation: Density of glycerine is more than that of water. Hence, glycerin exerts more upthrust on the block of wood than water, causing it to float in glycerine with a smaller volume.

**Question 16**

A body of volume V and density is kept completely immersed in a liquid of density . If g is the acceleration due to gravity, then write expressions for the following:

(i) The weight of the body, (ii) The upthrust on the body,

(iii) The apparent weight of the body in liquid, (iv) The loss in weight of the body.

**Answer 16**

(i) Weight of the body =

(ii) Upthrust on the body =

(iii) Apparent weight of the body in liquid =

(iv) Loss in weight of the body =

**Question 17**

A body held completely immersed inside a liquid experiences two forces:

(i) F_{1}, the force due to gravity and

(ii) F_{2}, the buoyant force.

Draw a diagram showing the direction of these forces acting on the body and state the condition when the body will float or sink.

**Answer 17**

If F1 F2 or F1 = F2, the body will float.

If F1 F2, the body will sink.

**Question 18**

Complete the following sentences:

(a) Two balls, one of iron and the other of aluminium experience the same upthrust when dipped completely in water if _____________ .

(b) An empty tin container with its mouth closed has an average density equal to that of a liquid. The container is taken 2 m below the surface of that liquid and is left there. Then the container will ____________ .

(c) A piece of wood is held under water. The upthrust on it will be

___________ the weight of the wood piece.

**Answer 18**

(a) Both have equal volumes. (b) Bounce back to the surface.

(c) More than

**Question 19**

Prove that the loss in weight of a body when immersed wholly or partially in a liquid is equal to the buoyant force (or upthrust) and this loss is because of the difference in pressure exerted by liquid on the upper and lower surfaces of the submerged part of body.

**Answer 19**

Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density as shown in the figure above. Let the upper surface PQ of the body is at a depth h_{1} while its lower surface RS is at depth h_{2} below the free surface of liquid.

At depth h_{1}, the pressure on the upper surface PQ,

P_{1} = h_{1} g.

Therefore, the downward thrust on the upper surface PQ,

F_{1} = Pressure x Area = h_{1 } gA ……………….(i)

At depth h_{2}, pressure on the lower surface RS,

P_{2} = h_{2 } g

Therefore, the upward thrust on the lower surface RS,

F_{2} = Pressure x Area = h_{2} gA …………………(ii)

The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.

From the above equations (i) and (ii), it is clear that F_{2 } F_{1} because h_{2} h_{1} and therefore, body will experience a net upward force.

Resultant upward thrust or buoyant force on the body,

F_{B} = F_{2} F_{1}

= h_{2 } gA h_{1 } gA

= A (h_{2} h_{1}) g

However, A (h_{2} h_{1}) = V, the volume of the body is submerged in a liquid.

Therefore, upthrust F_{B} = V g.

Now, V g = Volume of solid immersed x Density of liquid x Acceleration due to gravity

= Volume of liquid displaced x Density of liquid x Acceleration due to gravity

and = Mass of liquid displaced x Acceleration due to gravity

hence = Weight of the liquid displaced by the submerged part of the body

Thus, Upthrust F_{B} = weight of the liquid displaced by the submerged part of the body…..

##### (iii)

Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.

Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.

When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.

From the diagram, it is clear that

Loss in weight (Weight in air – Weight in water) = Volume of water displaced.

Or, Loss in weight = Volume of water displaced x 1 gcm^{-3} [Because the density of water = 1 gcm^{-3}]

Or, Loss in weight = Weight of water displaced ……………(iv)

From equations (iii) and (iv),

Loss in weight = Upthrust or buoyant force

**Question 20**

A sphere of iron and another sphere of wood of the same radius are held under water. Compare the upthrust on the two spheres.

Hint: Both have equal volume inside the water.

**Answer 20**

Since the spheres have the same radius, both will have an equal volume inside water, and hence, the upthrust acted by water on both the spheres will be the same.

Hence, the required ratio of upthrust acting on two spheres is 1:1.

**Question 21**

A sphere of iron and another of wood, both of same radius are placed on the surface of water. State which of the two will sink? Give a reason for your answer.

**Answer 21**

Sphere of iron will sink.

Density of iron is more than the density of water, so the weight of iron sphere will be more than the upthrust due to water in it; thus, it causes the iron sphere to sink.

Density of wood is less than the density of water, so the weight of sphere of wood shall be less than the upthrust due to water in it. So, the sphere of wood will float with a volume submerged inside water which is balanced by the upthrust due to water.

**Question 22**

How does the density of material of a body determine whether it will float or sink in water?

**Answer 22**

The bodies of average density greater than that of the liquid sink in it. While the bodies of average density equal to or smaller than that of liquid float on it.

**Question 23**

A body of density is immersed in a liquid of density. State the condition when the body will (i) float and (ii) sink in the liquid.

**Answer 23**

(i) The body will float if or =

(ii) The body will sink if

**Question 24**

It is easier to lift a heavy stone under water than in air. Explain.

**Answer 24**

It is easier to lift a heavy stone under water than in air because in water, it experiences an upward buoyant force which balances the actual weight of the stone acting downwards. Thus, due to upthrust there is an apparent loss in the weight of the heavy stone, which makes it lighter in water, and hence easy to lift.

**Question 25**

State the Archimedes’ principle.

**Answer 25**

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 liquid displaced by it.

**Question 26**

Describe an experiment to verify the Archimedes’ principle.

**Answer 26**

Let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight (Fig a).

Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid gets collected in the measuring cylinder.

When water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.

From diagram, it is clear that

Loss in weight (Weight in air weight in water) = 300 gf 200 gf = 100 gf

Volume of water displaced = Volume of solid = 100 cm^{3}

Because density of water = 1 gcm^{-3 }

Weight of water displaced = 100 gf = Upthrust or loss in weight

This verifies Archimedes’ principle

**Upthrust in Fluids, Archimedes’ Principle and Floatation **** MCQ-5(A)**

**Question 1**

A body will experience minimum upthrust when it is completely immersed in which of the following liquids:

(a) Turpentine (b) Water

(c) Glycerine (d) Mercury

**Answer 1**

(a) Turpentine

**Question 2**

The S.I. unit of upthrust is given by the following unit:

(a) Pa (b) N

(c) kg (d) kg m^{2}

**Answer 2**

(b) N

**Question 3**

A body of density sinks in a liquid of density . The densities and are related as shown below:

(a) = (b)

(c) (d) Nothing can be said.

**Answer 3**

** Selina Concise NUMERICAL -5(A) ****Upthrust in Fluids, Archimedes’ Principle and Floatation**

**Question 1**

A body of volume 100 cm^{3} weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 x 10^{3} kg m^{-3}. Find:

(i) The upthrust due to liquid and

(ii) The weight of the body in liquid.

**Answer 1**

**Question 2**

A body weighs 450 gf in air and 310 gf when completely immersed in water. Find the following factors:

(i) The volume of the body,

(ii) The loss in weight of the body, and (iii) The upthrust on the body.

State the assumption made in part (i).

**Answer 2**

**Question 3**

You are provided with a hollow iron ball A of volume 15 cm^{3} and mass 12 g and a solid iron ball B of mass 12 g. Both are placed on the surface of water contained in a large tub.

(a) Find upthrust on each ball.

(b) Which ball will sink? Give a reason for your answer. (Density of iron = 8.0 g cm^{-3})

**Answer 3**

**Question 4**

A solid of density 5000 kg m^{-3} weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m^{-3}. Calculate the apparent weight of the solid in water.

**Answer 4**

**Question 5**

Two spheres A and B, each of volume 100 cm^{3} are placed on water (density = 1.0 g cm^{-3}). The sphere A is made of wood of density 0.3 g cm^{-3} and the sphere B is made of iron of density 8.9 g cm^{-3}.

(a) Find:

(i) The weight of each sphere, and

(ii) The upthrust on each sphere.

(b) Which sphere will float? Give reason.

**Answer 5**

**Question 6**

The mass of a block made of certain material is 13.5 kg and its volume is 15 × 10^{-3} m^{3}.

(a) Calculate upthrust on the block if it is held fully immersed in water.

(b) Will the block float or sink in water when released? Give a reason for your answer.

(c) What will be the upthrust on block while floating?

Take density of water = 1000 kg m^{-3}.

**Answer 6**

**Question 7**

A piece of brass weighs 175 gf in air and 150 gf when fully submerged in water. The density of water is 1.0 g cm^{3}.

(i) What is the volume of the brass piece? (ii) Why does the brass piece weigh less in water?

**Answer 7**

**Question 8**

A metal cube of edge 5 cm and density 9 g cm^{-3} is suspended by a thread so as to be completely immersed in a liquid of density 1.2 g cm^{-3}. Find the tension in thread. (Take g = 10 m s^{-2})

**Answer 8**

**Question 9**

A block of wood is floating on water with its dimensions 50 cm x 50 cm x 50 cm inside water. Calculate the buoyant force acting on the block. Take g = 9.8 N kg^{-1}.

**Answer 9**

**Question 10**

A body of mass 3.5 kg displaces 1000 cm^{3} of water when fully immersed inside it. Calculate: (i) the volume of body, (ii) the upthrust on body and (iii) the apparent weight of body in water.

**Answer 10**

**ICSE Class 9th Exe- 5(B) ****Upthrust in Fluids, Archimedes’ Principle and Floatation**

**Question 1**

Define the term density.

**Answer 1**

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

**Question 2**

What are the unit of density in (i) C.G.S (ii) S.I system

**Answer 2**

(i) The C.G.S. unit of density is gcm^{-3}.

(ii) The S.I. unit of density is kgm^{-3}.

**Question 3**

Express the relationship between the C.G.S. and S.I. units of density.

**Answer 3**

1 gcm^{-3} = 1000 kgm^{-3}

**Question 4**

‘The density of iron is 7800 kg m^{-3}‘. What do you understand by this statement?

**Answer 4**

It means the mass of 1 m^{-3} of iron is 7800 kg.

**Question 5**

Write the density of water at in S.I. units.

**Answer 5**

Density of water at in S.I. units is 1000 kgm^{-3}.

**Question 6**

How are the (i) Mass, (ii) Volume and (iii) Density of a metallic piece affected, if at all, with an increase in temperature?

**Answer 6**

(i) Mass of a metallic body remains unchanged with increase in temperature.

(ii) Volume of metallic body increases with an increase in temperature.

(iii) Density (= Mass/volume) of a metallic body decreases with an increase in temperature.

**Question 7**

Water is heated from to. How does the density of water change with temperature?

**Answer 7**

On heating from the density of water increases up to and then decreases beyond

**Question 8**

Complete the following sentences.

(i) Mass = ……………….. × density

(ii) S.I. unit of density is ………. .

(iii) Density of water is …. … kg m^{-3}.

(iv) Density in kg m^{-3} = ………… × density in g cm^{-3}

**Answer 8**

(i) Volume, (ii) kg m^{-3}, (iii) 1000 and (iv) 1000

**Question 9**

What do you understand by the term relative density of a substance?

**Answer 9**

The relative density of a substance is the ratio of density of that substance to the density of water at

**Question 10**

What is the unit of relative density?

**Answer 10**

Relative density is the ratio of two similar quantities; thus, it has no unit.

**Question 11**

Differentiate between density and relative density of a substance.

**Answer 11**

Density of a substance is the ratio of its mass to its volume but R.D. of a substance is the ratio of density of that substance to the density of water at

**Question 12**

With the use of Archimedes’ principle, state how you will find relative density of a solid denser than water. How will you modify your experiments if the solid is soluble in water?

**Answer 12**

Steps:

(i) With the help of a physical balance, find the weight, W_{1} of the given solid.

(ii) Immerse the solid completely in a beaker filled with water such that it does not touch the walls and bottom of beaker, and find the weight W_{2} of solid in water.

Observations:

Loss in weight of solid when immersed in water = (W_{1} W_{2}) gf

R.D. = Weight of solid in air/Loss of weight of solid in water

R.D. = W_{1}/(W_{1} W_{2}).

If the solid is soluble in water, then instead of water, take a liquid in which the solid is insoluble and it sinks in the liquid.

Then, R.D. = (Weight of solid in air/Loss of weight of solid in liquid) x R.D. of the liquid

**Question 13**

A body weighs W gf in air and W_{1} gf when it is completely immersed in water. Find: (i) Volume of the body, (ii) Upthrust on the body and (iii) Relative density of the material of the body.

**Answer 13**

**Question 14**

Describe an experiment, using Archimedes principle, to find relative density of a liquid.

**Answer 14**

Relative density is the ratio of weight of a given volume of liquid to the weight of the same volume of water.

Using Archimedes principle, we can perform an experiment which measures the weight of a liquid displaced by a body and weight of water displaced by the same body.

Weight of liquid displaced by a body is given by the difference of weight of a body in air and weight of a body in liquid.

Weight of the water displaced by the body can be found by knowing the difference of the weight of the body in air and the weight of the body in water.

Therefore, using Archimedes principle, the relative density can be calculated using the formula:

**Question 15**

A body weighs W_{1}gf in air and when immersed in a liquid it weighs W_{2}gf, while it weights W_{3}gf on immersing it in water. Find: (i) volume of the body (ii) upthrust due to liquid (iii) relative density of the solid and (iv) relative density of the liquid.

**Answer 15**

(i) Volume of the body = W_{1 }– W_{3 }cm^{3}

(ii) Upthrust due to liquid = loss in weight when immersed in liquid = W_{1} – W_{2 }gf

(iii)

Weight of a body in air = W_{1}gf

and Weight of that body in liquid = W_{2}gf

so Weight of that body in water = W_{3}gf

(iv)

Weight of a body in air = W_{1}gf

and Weight of that body in liquid = W_{2}gf

so Weight of that body in water = W_{3}gf

** MCQ-5(B) ****Upthrust in Fluids, Archimedes’ Principle and Floatation Selina Physics Solution**

**Question 1**

Relative density of a substance is expressed by comparing the density of that substance with the density of :

(a) air (b) mercury (c) water (d) iron

**Answer 1**

(c) water

**Question 2**

The unit of relative density is :

(a) g cm^{-3 }(b) kg m^{-3} (c) m^{3} kg^{-1} (d) no unit

**Answer 2**

No unit.

**Question 3**

The density of water is as listed below:

(a) 1000 g cm^{-3} (b) 1 kg m^{-3} (c) 1 g cm^{-3} (d) None of these.

**Answer 3**

1 g cm^{-3}

**NUMERICAL 5(B) ****Upthrust in Fluids, Archimedes’ Principle and Floatation Revised ICSE Physics 9th**

**Question 1**

The density of copper is 8.83 g cm^{-3}. Express it in kg m^{-3}.

**Answer 1**

**Question 2**

The relative density of mercury is 13.6. State its density in (i) C.G.S. unit and (ii) S.I. unit.

**Answer 2**

**Question 3**

The density of iron is 7.8 x 10^{3} kg m^{-3}. What is its relative density?

**Answer 3**

**Question 4**

The relative density of silver is 10.8. Find its density.

**Answer 4**

**Question 5**

Calculate the mass of a body whose volume is 2 m^{3} and relative density is 0.52.

**Answer 5**

**Question 6**

Calculate the mass of air in a room of dimensions 4.5 m × 3.5 m × 2.5 m if the density of air at N.T.P. is 1.3 kgm^{-3 .}

**Answer 6**

**Question 7**

A piece of stone of mass 113 g sinks to the bottom in water contained in a measuring cylinder and water level in cylinder rises from 30 ml to 40 ml. Calculate R.D. of stone.

**Answer 7**

**Question 8**

A body of volume 100 cm^{3} weighs 1 kgf in air. Find: (i) Its weight in water and (ii) Its relative density.

**Answer 8**

**Question 9**

A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm^{3} of water. Find: (i) The mass of body in water and (ii) The relative density of material of the body.

**Answer 9**

**Question 10**

A solid weighs 120 gf in air and and 105 gf when it is completely immersed in water. Calculate the relative density of solid.

**Answer 10**

**Question 11**

A solid weighs 32 gf in air and 28.8 gf in water. Find: (i) The volume of solid, (ii) R.D. of solid and (iii) The weight of solid in a liquid of density 0.9 g cm^{-3}.

**Answer 11**

**Question 12**

A body weighs 20 gf in air and 18.0 gf in water. Calculate the relative density of the material of the body.

**Answer 12**

**Question 13**

A solid weighs 1.5 kgf in air and 0.9 kgf in a liquid of density 1.2 x 10^{3} kg m^{-3}. Calculate R.D. of solid.

**Answer 13**

**Question 14**

A jeweler claims that he makes ornaments of pure gold that has a relative density of 19.3. He sells a bangle weighing 25.25 gf to a person. The clever customer weighs the bangle when immersed in water and finds that it weighs 23.075 gf in water. With the help of suitable calculations, find out whether the ornament is made of pure gold or not.

Hint : Calculate R.D. of the material of the bangle.

**Answer 14**

**Question 15**

A piece of iron weighs 44.5 gf in air. If the density of iron is 8.9 × 10^{3}, find the weight of iron piece when immersed in water.

**Answer 15**

Density of iron = 8.9 × 10^{3} = 8900

Density of water = 1000

Weight of iron when immersed in water is given by

**Question 16**

A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:

(a) The weight of the piece of stone in air,

(b) The volume of the piece of stone,

(c) The relative density of stone,

(d) The relative density of the liquid.

**Answer 16**

(a) The mass of stone is 15.1 g. Hence, its weight in air will be W_{a} = 15.1 gf

(b) When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 – 9.7 = 5.4 gf, Since the density of water is 1 g cm^{-3}, the volume of stone is 5.4 cm^{3}.

(c) Weight of stone in liquid is W_{l} = 10.9 g

- Weight of stone in water is Ww = 9.7 gf
- Therefore, the relative density of stone is

**EXE – 5(C) Chapter-5 ****Upthrust in Fluids, Archimedes’ Principle and Floatation**

**Question 1**

State the principle of floatation.

**Answer 1**

According to the principle of floatation, the weight of a floating body is equal to the weight of the liquid displaced by its submerged part.

**Question 2 Upthrust Archimedes Principle Floatation**

A body is held immersed in a liquid. (i) Name the two forces acting on the body and draw a diagram to show these forces. (ii) State how the magnitudes of two forces mentioned in part (i) determine whether the body will float or sink in liquid when it is released. (iii) What is the net force on the body if it (a) sinks and (b) floats?

**Answer 2**

(i) Two forces acting on the body are as listed below:

(a) Weight of the body (downwards)

(b) Upthrust of the liquid (upwards)

(ii) If the weight of the body is greater than the upthrust acting on it, the body will sink

If the weight of the body is equal to or less than the upthrust acting on it, the body will float.

(iii) (a) The net force acting on the body when it sinks is body’s own weight.

(b) The net force acting on the body when it floats is the upthrust due to the liquid.

**Question 3**

When a piece of wood is suspended from the hook of a spring balance, it reads 70 gf. The wood is now lowered into water. What reading do you expect on the scale of spring balance?

**Answer 3**

The reading on the spring balance will be zero because wood floats on water and while floating the apparent weight = 0.

**Question 4**

A solid iron ball of mass 500 g is dropped in mercury contained in a beaker. (a) Will the ball float or sink? (b) What will be the apparent weight of the ball? Give reasons

**Answer 4**

(a) The ball will float because the density of ball (i.e. iron) is less than the density of mercury.

(b) While floating, the apparent weight = 0.

**Question 5**

How does the density of a substance determine whether a solid piece of that substance will float or sink in a given liquid?

**Answer 5**

The body will float if its density is less than or equal to the density of the liquid.

The body will sink if its density is greater than the density of the liquid.

**Question 6**

Explain why an iron nail floats on mercury, but it sinks in water.

Hint: Density of iron is less than that of mercury, but more than that of water.

**Answer 6**

Density of iron is less than the density of mercury; hence, an iron nail floats in mercury and density of iron is more than the density of water; hence, an iron nail sinks in water.

**Question 7**

Explain why an iron nail floats on mercury, but it sinks in water.

Hint: Density of iron is less than that of mercury, but more than that of water.

**Answer 7**

Density of iron is less than the density of mercury; hence, an iron nail floats in mercury and density of iron is more than the density of water; hence, an iron nail sinks in water.

**Question 8**

A homogeneous block floats on water (a) partly immersed (b) completely immersed. In each case state the position of center of buoyancy B with respect to the center of gravity G of the block.

**Answer 8**

When the body is partially immersed, its center of buoyancy will be below the center of gravity of the block.

When the body is completely immersed, its center of buoyancy will coincide the center of gravity.

**Question 9**

Fig. 5.15 shows the same block of wood floating in three different liquids A, B and C of densities, and respectively. Which of the liquid has the highest density? Give a reason for your answer.

**Answer 9**

The upthrust on the body by each liquid is the same and equal to the weight of the body.

However, upthrust = Volume submerged × × g,

For the liquid C, since the volume submerged is least so the density ρ_{3} must be maximum.

**Question 10**

Draw a diagram to show the forces acting on a body floating in water with its some part submerged. Name the forces and show their points of application. How is the weight of water displaced by the floating body related to the weight of the body itself?

**Answer 10**

The forces acting are as listed below:

(i) Weight of the body acting downwards.

(ii) Upthrust due to water acting upwards.

Weight of water displaced by the floating body = Weight of the body

**Question 11**

What is the centre of buoyancy? State its position for a floating body with respect to the centre of gravity of the body.

**Answer 11**

Centre of buoyancy: It is the point through which the resultant of the buoyancy forces on a submerged body act; it coincides with the centre of gravity of the displaced liquid, if the body is completely immersed.

For a floating body with its part submerged in the liquid, the centre of buoyancy is at the centre of gravity of the submerged part of the body and it lies vertically below the centre of gravity of the entire body.

**Question 12**

A balloon filled with helium gas floats in a big closed jar which is connected to an evacuating pump. What will be your observation, if air from the jar is pumped out? Explain your answer.

**Answer 12**

**Observation **: The balloon will sink.

**Explanation **: As air is pumped out from jar, the density of air in jar decreases, so the upthrust on balloon decreases. As weight of balloon exceeds the upthrust on it, it sinks.

**Question 13**

A block of wood is so loaded that it just floats in water at room temperature. What change will occur in the state of floatation, if

(a) Some salt is added to water, (b) Water is heated?

Give reasons.

**Answer 13**

(a) It will float with some part outside water.

**Reason **: On adding some salt to water, the density of water increases, so upthrust on a block of wood increases, and hence, the block rises up till the weight of salty water displaced by the submerged part of block becomes equal to the weight of the block.

(b) The block will sink.

**Reason: **On heating, the density of water decreases, so upthrust on the block decreases and the weight of block exceeds upthrust due to which it sinks.

**Question 14**

**Answer 14**

**Question 15**

Two identical pieces, one of ice (density = 900kg per meter cube) and other wood (density = 300kg per meter cube) float on water.

(a) Which of the two will have more volume submerged inside water

(b) Which of two will experience more upthrust due to water.

**Answer 15**

(a) Ice will be more submerged inside water. Ice has a greater density than wood, although the volume of both is the same. So to support a greater amount of weight, ice needs to displace more water, and to displace more water, it has to be submerged more as compared to wood.

(b) As ice displaces more water, it will experience more upthrust.

**Question 16**

Why is the floating ice less submerged in brine than in water?

**Answer 16**

Density of brine is more than the density of water. Hence, the upthrust exerted by brine is more than the upthrust exerted by water on ice. Therefore, floating ice is less submerged in brine.

**Question 17**

A man first swims in sea water and then in river water. (i) Compare the weights of sea water and river water displaced by him.

(ii) Where does he find it easier to swim and why?

**Answer 17**

(i) 1:1; The weight of the water displaced by the man in sea and river will be same and will be equal to his own weight.

(ii) He finds it easier to swim in the sea because the density of sea water is more than the density of river water. So his weight is balanced in sea water with a part of his body submerged in the water.

**Question 18**

An iron nail sinks in the water while an iron ship floats on water. Explain the reason.

**Answer 18**

An iron nail sinks in water because density of iron is more than the density of water, so the weight of the nail is more than the upthrust of water on it.

On the other hand, ships are also made of iron, but they do not sink. This is because the ship is hollow and the empty space in it contains air, which makes its average density less than that of water. Therefore, even with a small portion of ship submerged in water, the weight of water displaced by the submerged part of ship becomes equal to the total weight of ship and it floats.

**Question 19**

What can you say about the average density of a ship floating on water in relation to the density of water?

**Answer 19**

Due to the hollow and empty space in the ship, the average density of a ship is less than the density of water.

**Question 20**

A piece of ice floating in a glass of water melts, but the level of water in the glass does not change.

Give reasons.

**Hint: **Ice contracts on melting.

**Answer 20**

When a floating piece of ice melts into water, it contracts by the volume equal to the volume of ice pieces above the water surface while floating on it. Hence, the level of water does not change when ice floating on it melts.

**Question 21**

A buoy is held inside water contained in a vessel by tying it with a thread to the base of the vessel. Name the three* *forces that keep the body in equilibrium, and state the direction in which each force acts.

**Answer 21**

Forces acting on the body are listed below:

(i) Weight of the body vertically downwards.

(ii) Upthrust of water on body vertically upwards.

(iii) Tension in thread vertically downwards.

**Question 22**

A loaded cargo ship sails from sea water to river water. List your observations.

**Answer 22**

A ship submerges more as it sails from sea water to river water.

Density of river water is less than the density of sea water. Hence, according to the law of floatation, to balance the weight of the ship, a greater volume of water is required to be displaced in river water of lower density.

**Question 23 Upthrust Archimedes Principle Floatation**

Explain the following observations listed below:

(a) Icebergs floating in sea are dangerous for ships.

(b) An egg sinks in fresh water, but floats in a strong salt solution.

(c) Toy balloons filled with hydrogen rise to the ceiling, but if they are filled with carbon dioxide, then they sink to the floor.

(d) As a ship in harbour is being unloaded, it slowly rises higher in water.

(e) A balloon filled with hydrogen rises to a certain height and then stops rising further.

(f) A ship submerges more as it sails from sea water to river water.

#### Answer 23

##### (a) Icebergs are dangerous for ships

as they may collide with them. Icebergs being lighter than water, float on water with a major part of their surfaces laying under the water surface and only a small part lies outside water. Thus, it becomes difficult for the driver of the ship to estimate the size of the iceberg.

##### (b) Density of a strong salt solution is more than the density of fresh water

. Hence, the salt solution exerts a greater upthrust on the egg which balances the weight of the egg, so the egg floats in a strong salt solution but sinks in fresh water.

##### (c) Density of hydrogen is much less than the density of carbon dioxide.

When a balloon is filled with hydrogen, the weight of the air displaced by an inflated balloon (i.e. upthrust) becomes more than the weight of a gas filled balloon, and hence, it rises. In case of a balloon filled with carbon dioxide, weight of the balloon becomes more than the upthrust of the air, and hence, it sinks to the floor.

##### (d) As a ship in harbor is unloaded, its weight decreases.

As a result, it displaces less water, and the ship’s hull rises in water till the weight of the water displaced balances the weight of the unloaded ship.

##### (e) The reason is that the density of air decreases with altitude.

Therefore, as the balloon gradually goes up, the weight of the displaced air (i.e. uphrust) decreases. It keeps on rising as long as the upthrust exceeds its weight. When upthrust becomes equal to its weight, it stops rising.

##### (f) Density of river water is less than the density of sea water.

Hence, according to the law of floatation, to balance the weight of the ship, a great volume of water is required to be displaced in river water having a comparitively lower density.

**MCQ- 5(C) ****Upthrust in Fluids, Archimedes’ Principle and Floatation Selina solution**

**Question 1**

For a floating body, its weight W and upthrust F_{B} on it are related as listed below:

(a) W F_{B }(b) W F_{B}

(c)W= F_{B }(d)Nothing can be said.

**Answer 1**

W= F_{B}

**Question 2**

A body of weight W is floating in a liquid. Its apparent weight will be observed as listed below:

(a) Equal to W (b) Less than W

(c) Greater than W (d) Zero

**Answer 2**

Zero

**Question 3**

A body floats in a liquid A of density with a part of it submerged inside the liquid, while in liquid B of density _{ }it is totally submerged inside the liquid. The densities and are related as shown below:

(a) = (b) _{ }

(c) (d) Nothing can be said

**Answer 3**

_{ }

**NUMERICAL- 5(C) ****Upthrust in Fluids, Archimedes’ Principle and Floatation ICSE Class 9th Physics**

**Question 1**

A rubber ball floats on water with its 1/3^{rd} volume outside water. What is the density of rubber?

**Answer 1**

**Question 2**

A block of wood of mass 24 kg floats in water. The volume of wood is 0.032 m^{3}. Find the following factors listed below:

(a) The volume of block below the surface of water, (b) The density of wood.

(Density of water = 1000 kg m^{-3})

**Answer 2**

**Question 3 Upthrust Archimedes Principle Floatation**

A wooden cube of side 10 cm has mass 700 g. What part of it remains above the water surface while floating vertically on water surface?

**Answer 3**

**Question 4**

A piece of wax floats in brine. What fraction of its volume will be immersed?

Density of wax = 0.95 g cm^{-3}, Density of brine = 1.1 g cm^{-3}.

**Answer 4**

**Question 5**

If the density of ice is 0.9 g cm^{-3}, then what portion of an iceberg will remain below the surface of water in sea? (Density of sea water = 1.1 g cm^{-3})

**Answer 5**

**Question 6**

A piece of wood of uniform cross section and height 15 cm floats vertically with its height 10 cm in water and 12 cm in spirit. Find the densities of wood and spirit.

**Answer 6**

**Question 7**

A wooden block floats in water with two-third of its volume submerged. (a) Calculate the density of wood. (b) When the same block is placed in oil, three-quarters of its volume is immersed in oil. Calculate the density of oil.

**Answer 7**

**Question 8**

The density of ice is 0.92 g cm^{-3} and that of sea water is 1.025 g cm^{-3}. Find the total volume of an iceberg which floats with its volume 800 cm^{3} above water.

**Answer 8**

**Question 9**

A weather forecasting plastic balloon of volume 15 m^{3} contains hydrogen of density 0.09 kg m^{-3}. The volume of an equipment carried by the balloon is negligible compared to its own volume. The mass of an empty balloon alone is 7.15 kg. The balloon is floating in air of density 1.3 kg m^{-3}.

Calculate: (i) The mass of hydrogen in the balloon, (ii) The mass of hydrogen and balloon, (iii) The total mass of hydrogen, balloon and equipment if the mass of equipment is *x *kg, (iv) The mass of air displaced by balloon and (v) The mass of equipment using the law of floatation.

**Answer 9**

##### –: End of Upthrust Archimedes Principle Floatation :–

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