**Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5B** Relative Density and it’s Measurement by 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 session**. **Visit official Website CISCE for detail information about ICSE Board Class-9.

**Upthrust in Fluids, Archimedes’ Principle and Floatation Exe-5B 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 – 5B | Relative Density and it’s Measurement by Archimedes’ Principle |

Topics | Solution of Exe-5(B) Numericals Answer Type |

Academic Session | 2023-2024 |

**Exe-5B Relative Density and it’s Measurement Numericals Answer Type**

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

**Page 129**

**Question 1. **The density of copper is 8.83 g cm^{-3}. Express it in kg m^{-3}.

**Answer:**

1 Density of copper in C.G.S. = 8.83 gcm^{-3}

Density of copper in S.I. = 8.83/1000×100^{-3}

= 8830 kgm^{-3}

**Question 2. **The relative density of mercury is 13.6. State its density in (i) C.G.S. unit and (ii) S.I. unit.

**Answer:**

R.D. of mercury = 13.6

(i) Density in C.G.S. = 13.6 gcm^{-3}

(ii) Density in S.I. = 13.6 × 10^{3} kgm^{-3}

**Question 3. **The density of iron is 7.8 x 10^{3} kg m^{-3}. What is its relative density?

**Answer:**

Density of iron = 7.8×10^{3} kgm^{-3}

Density of iron in C.G.S = 7.8 gcm^{-3}

R.D. = Density of C.G.S. (without unit) = 7.8

**Question 4. **The relative density of silver is 10.8. Find its density.

**Answer:**

R.D. of silver = 10.8

Density of silver in C.G.S. = 10.8 gcm^{-3}

Density in S.I. = 10.8 × 10^{3} kgm^{3}

**Question 5. **Calculate the mass of a body whose volume is 2 m^{3} and relative density is 0.52.

**Answer:**

R.D. of silver = 0.52

Volume = 2 m^{3}

Density of body in S.I. = 0.52 × 10^{3} kgm^{3}

∴ Mass = Density × volume = (0.52 × 10^{3}) × 2 = 1040 kg

**(Upthrust in Fluids Exe-5B Numericals ICSE)**

**Question 6. **Calculate the mass of air in a room of dimensions 4.5 m × 3.5 m × 2.5 m if the density of air at N.T.P. is 1.3 kgm^{-3 .}

**Answer:**

Volume of air = 4.5×3.5×2.5 m^{3}

Density pf air at NTP = 1.3 kgm^{3}

Mass of air = Density × volume

Or Mass = (1.3) × (4.5 × 3.5 × 2.5) = 51.19 kg

**Question 7. **A piece of stone of mass 113 g sinks to the bottom in water contained in a measuring cylinder and water level in cylinder rises from 30 ml to 40 ml. Calculate R.D. of stone.

**Answer:**

Mass of stone = 113 g

Rise in water level = (40 – 30) ml = 10 ml

This rise is equal to the space occupied (volume) by the stone .

∴ volume of stone = 10 cm^{-3}

Density of stone in C.G.S. = Mass/Volume = 113/10

=11.3 gcm^{-3}

R.D. = 11.3

**Question 8. **A body of volume 100 cm^{3} weighs 1 kgf in air. Find: (i) Its weight in water and (ii) Its relative density.

**Answer:**

Volume of body = 100 cm^{3}

Weight in air , W_{1} = 1 kgf = 1000 gf

Mass of body = 1 kg = 1000 g

R.D. of solid = 10

R.D. of water = 1

(i) Let W_{2} be the weight of the body in water.

R.D. of body = R.D. of waterW_{1}/W_{1}-W_{2 }× R.D. of water

or , 10 = 1000(1000- W_{2})×1

or , 10(1000 – W_{2}) = 1000

or , 1000 – W_{2} = 100

or , W_{2} = 900 gf

(ii) R.D. of body = Density in C.G .S. (without unit)

or , R.D. = Mass/Volume = 1000/100

= 10

**Question 9. **A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm^{3} of water. Find: (i) The mass of body in water and (ii) The relative density of material of the body.

**Answer:**

Weight of body = 70 kg

Volume of water displaced by body = 20,000 cm^{3} = 0.02 m^{3}

(i) Mass of solid immersed in water = mass of water displaced

Mass of solid immersed in water = Density of water × Volume of water displaced

Mass of solid immersed in water = 1000 kgm^{-3} × 0.02 m^{3} = 20 kg

(ii) R.D. of solid = Density in C.G.S. (without unit)

Density in C.G.S. = mass/volume =70 × 1000/20,000

=3.5 gcm^{3}

R.D. = 3.5

**Question 10. **A solid weighs 120 gf in air and and 105 gf when it is completely immersed in water. Calculate the relative density of solid.

**Answer:**

Weight of solid in air , W_{1} = 120 gf

Weight of solid when completely immersed in water W_{2} = 105 gf

R.D. of solid = R.D. of waterW_{1}/W_{1}-W_{2 }× R.D. of water

R.D. of solid = 120/120-105×1

R.D. of solid = 8

**(Upthrust in Fluids Exe-5B Numericals ICSE)**

**Question 11. **A solid weighs 32 gf in air and 28.8 gf in water. Find: (i) The volume of solid, (ii) R.D. of solid and (iii) The weight of solid in a liquid of density 0.9 g cm^{-3}.

**Answer:**

Weight of solid in air , W_{1} = 32 gf

Weight of solid when completely immersed in water W_{2} = 28.8 gf

(i) Volume of solid = Mass / density of solid

= 32/10 = 3.2 m^{3}

(ii) R.D. of solid = W_{1}/(W_{1}-W_{2}) x R.D. of water

R.D. of solid = 32/(32-28.8)×1

R.D. of solid = 10

( iii) Weight of solid in liquid fo density 0.9 gcm^{-3} = W_{3}

R.D. of solid = = W_{1}/(W_{1}-W_{3}) x R.D. of liquid

or , 10 = 32/(32-W_{3})×0.9

**or , W _{3} = 29.12 gf**

**Question 12. **A body weighs 20 gf in air and 18.0 gf in water. Calculate the relative density of the material of the body.

**Answer:**

Weight of body in air , W_{1} = 20 gf

Weight of body when completely immersed in water W_{2} = 18 gf

R.D. of body = W_{1}/(W_{1}-W_{2}) x R.D. of water

R.D. of body = 20/(20-18)×1

R.D. of body = 10

**Exe-5B Relative Density and it’s Measurement Numericals Answer Type**

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

**Page 130**

**Question 13. **A solid weighs 1.5 kgf in air and 0.9 kgf in a liquid of density 1.2 x 10^{3} kg m^{-3}. Calculate R.D. of solid.

**Answer:**

Weight of body in air , W_{1} = 1.5 kgf

Weight of body when completely immersed in liquid W_{2} = 0.9 kgf

Density of liquid = 1.2 × 10^{3 }kgm^{-3}

R.D. of liquid = 1.2

R.D. of body = W_{1}/(W_{1}-W_{2}) x R.D. of liquid

R.D. of body = 1.5/(1.5-0.9)×1.2

R.D. of body = (1.5/0.6)×1.2

R.D. of body = 3

**(Upthrust in Fluids Exe-5B Numericals ICSE)**

**Question 14. **A jeweler claims that he makes ornaments of pure gold that has a relative density of 19.3. He sells a bangle weighing 25.25 gf to a person. The clever customer weighs the bangle when immersed in water and finds that it weighs 23.075 gf in water. With the help of suitable calculations, find out whether the ornament is made of pure gold or not.

Hint : Calculate R.D. of the material of the bangle.

**Answer:**

R.D. of pure gold = 19.3

Weight of bangle in air, W_{1} = 25.25 gf

Weight of bangle when completely immersed in water W_{2} = 23.075 gf

R.D. of bangle = W_{1}/(W_{1}-W_{2}) x R.D. of water

R.D. of bangle = 25.25/(25.25 – 23.075)×1

R.D. of bangle = 11.6

The bangle is not made of pure gold as its density is not 19.3.

**Question 15. **A piece of iron weighs 44.5 gf in air. If the density of iron is 8.9 × 10^{3}, find the weight of iron piece when immersed in water.

**Answer:**

Density of iron = 8.9 × 10^{3} = 8900

Density of water = 1000

Weight of iron when immersed in water is given by

**Weight of iron in water = Weight of iron in air × (1 – density of water/density of iron)**

= 44.5 gf × (1 – 1000/8900)

= 39.5 kgf

**Question 16. **A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:

(a) The weight of the piece of stone in air,

(b) The volume of the piece of stone,

(c) The relative density of stone,

(d) The relative density of the liquid.

**Answer:**

(a) The mass of stone is 15.1 g. Hence, its weight in air will be W_{a} = 15.1 gf

(b) When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 – 9.7 = 5.4 gf, Since the density of water is 1 g cm^{-3}, the volume of stone is 5.4 cm^{3}.

(c) Weight of stone in liquid is W_{l} = 10.9 g

1. Weight of stone in water is Ww = 9.7 gf

2. Therefore, the relative density of stone is

3.

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

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