Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5A Numericals Answer Physics Class-9 ICSE Selina Publishers

Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5A Upthrust and Archimedes Principle Numericals Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Numericals Answer type Questions of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5A Numericals Answer

(ICSE Class – 9 Physics Concise Selina Publishers)

 Board ICSE Class 9 Subject Physics Writer / Publication Concise selina Publishers Chapter-5 Upthrust in Fluids, Archimedes’ Principle and Floatation Exe – 5A Upthrust and Archimedes Principle Topics Solution of Exe-5(A) Numericals Answer Type Academic Session 2023-2024

Exe-5A Upthrust and Archimedes Principle Numericals Answer Type

Ch-5 Upthrust in Fluids, Archimedes’ Principle and Floatation Physics Class-9 ICSE Concise

Page 122

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.

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).

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

Exe-5A Upthrust and Archimedes Principle Numericals Answer Type

Ch-5 Upthrust in Fluids, Archimedes’ Principle and Floatation Physics Class-9 ICSE Concise

Page 123

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)

(Upthrust in Fluids Exe-5A Numericals ICSE)

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.

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.

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.

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?

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
(Upthrust in Fluids Exe-5A Numericals ICSE)

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)

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.

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.

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 in Fluids, Archimedes’ Principle and Floatation Exe-5A Numericals Answer Type  Solutions :–

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