**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 session**. **Visit 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 1/3^{rd} volume outside water. What is the density of rubber?

**Answer:**

**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:**

**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:**

**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:**

**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:**

**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:**

Let densities of water, wood and spirit are ρ, ρ_{w} and ρ_{s} 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

Vρ_{w}g = A×10×ρ×g

ρ_{w }= (A×10×ρ×g)/(15×A×g) = 0.67 g/cm^{3}

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

Vρ_{w}g = A×12×ρ_{s}×g

ρ_{s }= 0.83 g/cm^{3}

**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:**

**(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 cm^{3} above water.

**Answer:**

**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:**

Volume of plastic balloon= 15 m^{3 }

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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