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.

Board ICSE
Publications Selina Publication
Subject Physics
Class 9th
Chapter-5 Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5(A)
Book Name Concise
Topics Solution of Exe-5(A),  MCQ-5(A), Numericals-5(A), Exercise-5(B)  MCQ-5(B)  Numericals-5(B) and Exercise-5(C)   MCQ-5(C)  Numericals-5(C)
Academic Session 2021-2022

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.


–: Select Topics :–

Exercise-5(A)  MCQ-5(A)  Numericals-5(A)

Exercise-5(B)  MCQ-5(B)  Numericals-5(B)

Exercise-5(C)   MCQ-5(C)  Numericals-5(C)

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

Page 109

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.

causes of upthrust

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:

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.


Page 110

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 F1 in river water and F2 in sea water when dipped up to the same level. Which is more, F1 or F2? Give reason.

Answer 14

F2   F1; 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 ρ is kept completely immersed in a liquid of density ρL . 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 =  Vρg

(ii) Upthrust on the body =  VρLg

(iii) Apparent weight of the body in liquid = V(ρ – ρL)g

(iv) Loss in weight of the body =  VρLg

Question 17

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

(i) F1, the force due to gravity and

(ii) F2, 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

float and sink

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

cylinderical

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 h1 while its lower surface RS is at depth h2 below the free surface of liquid.

At depth h1, the pressure on the upper surface PQ,

P1 = h  g.

Therefore, the downward thrust on the upper surface PQ,

F1 = Pressure x Area = h1  gA ……………….(i)

At depth h2, pressure on the lower surface RS,

P2 = h2  g

Therefore, the upward thrust on the lower surface RS,

F2 = Pressure x Area = h2 ρ 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 > F1 because h2  > h1 and therefore, body will experience a net upward force.

Resultant upward thrust or buoyant force on the body,

FB = F2  F1

= h2ρ gA  h gA

= A (h2  h1)  g

However, A (h2  h1) = V, the volume of the body is submerged in a liquid.

Therefore, upthrust FB = V  g.

Now, Vρ  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 FB = 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.

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 ρ <  or = ρL

(ii) The body will sink if  ρ  > ρL

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.

measuring cylinder 2

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 cm3

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)

Page 110

Question 1

A body will experience minimum upthrust when it is completely immersed in :

(a) Turpentine  (b) Water

(c) Glycerine   (d) Mercury

Answer 1

(a) Turpentine

Question 2

The S.I. unit of upthrust is :

(a) Pa (b) N

(c) kg   (d) kg m2

Answer 2

(b) N

Question 3

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

(a)  ρ   =  ρL (b)  ρ < ρL

(c) ρ >  ρL       (d) Nothing can be said.  ρ

Answer 3

(c) ρ >  ρL


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

Pahe 110

Question 1

A body of volume 100 cm3 weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 x 103 kg m-3. Find:

(i) The upthrust due to liquid and

(ii) The weight of the body in liquid.

Answer 1

weight of the body in air = W = 5 kgf

Volume of Body = V = 100³ cm = 100 x 10⁻⁶ = 10⁻⁴m³

Density of liquid = d= 1.8 x 10³ kg/m³

(i) The upthrust due to liquid

Upthrust= Buoyant Force B=Vdg

= 10⁻⁴ x 1.8×10³ x g
= 0.18g N
= 0.18 kgf

(ii) The weight of the body in liquid: So weight of Body in Liquid=W-B
= 5 kgf – 0.18 kgf
= 4.82 kgf

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

Weight of the body in air, W = 450 gf = 450g dynes

Weight of the body in water, W/ = 310 gf = 310g dynes

Let, d be the density of the body, V be its volume. Let q be the density of water.

W = Vdg

450g = Vdg

Buoyant force, B = Vqg

B = Vg [q = 1 g/cm3]

Now, W/ = W – B

310g = 450g – Vg

V = 140 cm3

Loss in weight = Buoyant force = Vqg = 140 × 1 × g = 140g dynes = 140 gf

Again, upthrust = Buoyant force = 140 gf


Page 111

Question 3

You are provided with a hollow iron ball A of volume 15 cm3 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

Upthrust Archimedes Principle Floatation Numerical 3 ICSE 9th physics

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

mass of solid= 0.5 kgf

volume of solid = mass of solid / density of solid

= 0.5/5000

= 0.0001 m³

volume of water displaced = volume of solid = 0.0001 m³

mass of water displaced = density of water x volume of water displaced =

1000 x 0.0001 = 0.1 kgf

apparent weight = weight in air – weight of liquid displaced

= 0.5 kgf – 0.1 kgf

= 0.4 kgf

Question 5

Two spheres A and B, each of volume 100 cm3 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

Density of water = 1gcm-3

Density of sphere A = 0.3 gcm-3

Density of sphere B = 8.9 gcm-3

Volume of sphere A & B = 100 cm3

(a) (i)To find the weight of sphere A and B

Weight of sphere A = density of sphere A x volume of sphere x g

= 0.3 x 100 x g = 30gf

Weight of sphere B = density of sphere B x volume of sphere x g

= 8.9 x 100 x g = 890gf

(ii) To find upthrust on each sphere

Upthrust on sphere A = volume of sphere A x density of water x g

= 100 x 1 x g = 100gf

Upthrust on sphere B = volume of sphere B x density of water x g

= 100 x 1 x g = 100 gf

Upthrust acting on both the spheres is the same as the volume of spheres A and B inside water is the same

(b) Sphere A will float as the density of wood is lesser than that of water.

Question 6

The mass of a block made of certain material is 13.5 kg and its volume is 15 × 10-3 m3.

(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

Numerical 6 ICSE 9th physics

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 cm3.

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

Answer 7

Weight of piece of brass in air = 175 gf
Weight pf piece of brass when fully immersed in water = 150 gf
Density of water = 1g cm-3
(i) Volume of brass price = Loss in weight = 175 – 150 = 25 cm3
(ii) The brass of price weighs less in water due to upthrust

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

Given , side of the cube = 5 cm

∴ volume of the cube = 5 × 5 × 5 = 125 cm3

Mass of the cube = volume × density

= 125 × 9 = 1125 g

∴ weight of the cube = 1125 gf (downwards)

Upthrust on cube = weight of the liquid displaced

= volume of the cube × density of liquid × g

= 125 × 1.2 × g

= 150 gf (upwards)

Tension in thread = Net downward force

= Weight of cube – Upthrust on cube

= 1125 – 150 = 975 gf = 9.75 N

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

Volume of block of wood = 50 cm × 50 cm × 50 cm = 125000 cm3 = 0.125 m3

Given , g = 9.8 m/s2

Buoyant force = Vρg

= 0.125 × 1000 × 9.8 N

= 1225 N

Question 10

A body of mass 3.5 kg displaces 1000 cm3 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

Mass of body = 3.5 kg

Weight of the body = 3.5 kgf

Volume of water displaced when body is fully immersed = 1000 cm3

(i) Volume of body when fully immersed in liquid = Volume of water displaced

∴ Volume of body = 1000 cm3 or 0.001 m3

(ii) Upthrust on body = Volume of body × Density of water × g

= 0.001 × 1000 × g

= 1 kgf

(iii) Apparent weight = True weight – Upthrust

= (3.5 – 1)

= 2.5 kgf

–: End of Upthrust Archimedes Principle Floatation :–

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