# 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 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)

#### 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.

#### 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.

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.

(Upthrust in Fluids Exe-5C Numericals ICSE)

#### 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.

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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