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

Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5C Floatation 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-5C 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 – 5C Floatation
Topics Solution of Exe-5(C) Numericals Answer Type
Academic Session 2023-2024

Exe-5C Floatation Numericals Answer Type

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

Page 138

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

Answer:

ICSE 9th physics solution 5(C) Numerical 1

Question 2. A block of wood of mass 24 kg floats in water. The volume of wood is 0.032 m3. 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:

ICSE 9th physics solution exe 5(C) 2

Question 3. 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:

ICSE 9th physics solution exe 5(C) 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:

ICSE 9th physics solution exe 5(C) 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:

ICSE 9th physics solution exe 5(C) 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:

 Upthrust Archimedes Principle FloatationICSE 9th physics solution exe 5(C) 6

Let densities of water, wood and spirit are ρ, ρw and ρs respectively.

Let A be the area of the wooden block.

Total volume of the wooden block is, V = 15A

1. Mass of wood = Mass of water displaced by the wood

wg = A×10×ρ×g

ρ= (A×10×ρ×g)/(15×A×g) = 0.67 g/cm3

2. Mass of the wood = Mass of then spirit displaced

wg = A×12×ρs×g

ρ= 0.83 g/cm3

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:

ICSE 9th physics solution exe 5(C) 7

(Upthrust in Fluids Exe-5C Numericals ICSE)

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 cm3 above water.

Answer:

ICSE 9th physics solution exe 5(C) 8

Question 9. A weather forecasting plastic balloon of volume 15 m3 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 kg, (iv) The mass of air displaced by balloon and (v) The mass of equipment using the law of floatation.

Answer:

Volume of plastic balloon= 15 m

Mass of empty balloon = 7.15 kg

Density of hydrogen= 0.09 kgm-3

Density of air= 1.3 kgm-3

(i)  Mass of hydrogen in the balloon= Volume of balloon x Density of hydrogen
Mass of hydrogen in the balloon= (15 x 0.09)kg= 1.35kg

(ii)  Mass of hydrogen and balloon = Mass of empty balloon + Mass of hydrogen in the balloon
Mass of hydrogen balloon = [7.15 + 1.35 ) kg = 8.5 kg

(iii) Given mass of equipment= x
Total mass of hydrogen, balloon and equipmemt = (8.5 + x) kg

(iv) Weight of air displaced by the balloon = upthrust = Volume of balloon x density of air x g
Mass of air displaced= Volume of balloon x density of air
= 15 x 1.3=19.5 kg

(v) Using the law of floatation,
Mass of air displaced= Total mass of hydrogen, balloon and equipmemt
or, 19.5 = 8.5 + x
or, x=11 kg
Thus, mass of the equipment 11 kg

—  : End of Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5C Numericals Answer Type  Solutions :–

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