ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-17.3 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-17 | Mensuration |

Writer / Book | Understanding |

Topics | Solutions of Exe-17.3 |

Academic Session | 2024-2025 |

**Mensuration Exe-17.3**

ML Aggarwal Class 10 ICSE Maths Solutions

**Question 1. ****Find the surface area of a sphere of radius ****14 cm**

**Answer :**

##### Given radius of the sphere, r = 14 cm

Surface area of the sphere = 4πr^{2}

= 4 × (22/7) × 14^{2}

= 4 × 22 × 14 × 2

= 2464 cm^{3}

Hence the surface area of the sphere is 2464 cm^{2}.

**Question 2. ****Find the surface area of a sphere of diameter 21 cm **

**Answer :**

Given diameter of the sphere, d = 21 cm

Radius, r = d/2 = 21/2 = 10.5

Surface area of the sphere = 4πr^{2}

= 4 × (22/7) × 10.5^{2}

= 1386 cm^{2}

Hence the surface area of the sphere is 1386 cm^{2}.

**Question 3. ****A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm**^{3}, find the mass of the shot-put.

^{3}, find the mass of the shot-put.

**Answer :**

Radius of the metallic shot-put = 4.9 cm

Volume of the sphere, V = (4/3)πr^{3}

V = (4/3) × (22/7) × 4.9^{3}

⇒ V = 493.005

⇒ V = 493 cm^{3} **(approx)**

Given Density = 7.8 g per cm^{3}

Density = Mass/Volume

Mass = Density × Volume

= 7.8 × 493

**= **3845.4 g

Hence, the mass of the shot put is 3845.4 g.

**Question 4. ****Find the diameter of a sphere whose surface area is 154 cm**^{2}.

^{2}.

**Answer :**

Surface area of a sphere = 154 cm^{2}

Surface area of the sphere = 4πr^{2}

4 × (22/7) × r^{2} = 154

⇒ r^{2} = 154 × 7/(22 × 4) = 49/4

⇒ r = √49/2

Diameter = 2 × r = 2 × √49/2 = √49 = 7

Hence the diameter of the sphere is 7 cm.

**Mensuration Exe-17.3**

ML Aggarwal Class 10 ICSE Maths Solutions

Page 413

**Question 5. ****Find:**

**(i) the curved surface area.**

**(ii) the total surface area of a hemisphere of radius 21 cm.**

**Answer :**

Radius of a hemisphere = 21 cm

**(i) Curved surface area = 2πr**^{2}

^{2}

Curved surface area of the hemisphere = 2πr^{2}

= 2 × (22/7) × 21^{2}

= 2 × 22 × 3 × 21

= 2772 cm^{2}

Hence the curved surface area of the hemisphere is 2772 cm^{2}.

**(ii)** Total surface area of the hemisphere = 3πr^{2}

= 3 × (22/7) × 21^{2}

= 3 × 22 × 3 × 21

= 4158 cm^{2}

Hence, the total surface area of the hemisphere is 4158 cm^{2}.

**Question 6. ****A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm**^{2}.

^{2}.

**Answer :**

The inner diameter of hemispherical bowl = 10.5 cm

Radius, r = d/2 = 10.5/2 = 5.25 cm

Curved surface area of the bowl = 2πr^{2}

= 2 × (22/7) × 5.25^{2}

= 173.25 cm^{2}

Rate of tin plating = Rs.16 per 100 cm^{2}

So total cost = 173.25 × 16/100 = 27.72

Hence the cost of tin plating the bowl on the inside is Rs. 27.72.

**Question 7. ****The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.**

**Answer :**

Original radius of balloon = 7 cm

Radius after filling the air in it = 14 cm

The surface area of balloon, the original position

Surface area of the sphere = 4πr^{2}

Ratio of surface areas of the balloons = 4πr^{2}/4πR^{2}

= r^{2}/R^{2}

= 7^{2}/14^{2}

= 1/4

Hence the ratio of the surface areas of the spheres is 1:4.

**Question 8. ****A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π**

**Answer :**

Let the edge of a cube = a

Surface area = 6a^{2}

and surface area of sphere = 6a^{2}

Given sphere and cube has same surface area.

4πr^{2} = 6a^{2}

⇒ πr^{2}/a^{2} = 6/4

∴ r/a = √6/2√π

Volume of the sphere, V_{1 }= (4/3)πr^{3}

Volume of the cube, V_{2 }= a^{3}

∴ V_{1}/V_{2} = 4πr^{3}/3a^{3}

⇒ V_{1}/V_{2} = 4π√6^{3}/(3 × 2√π)^{3}

⇒ V_{1}/V_{2} = (4π × 6√6)/(3 × 8π × √π)

∴ V_{1}/V_{2} = √6/√π

Hence proved.

**Question 9.**

**(a) If the ratio of the radii of two sphere is 3 : 7, find :**

**(i) the ratio of their volumes.**

**(ii) the ratio of their surface areas.**

**(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.**

**Answer :**

**(a) Ratio in radii of two spheres = 3 : 7**

Volume of sphere = (4/3)πr^{3}

Ratio of the volumes = (4/3)πr_{1}^{3}/(4/3)πr_{2}^{3}

= r_{1}^{3}/ r_{2}^{3}

= 3^{3}/7^{3}

= 27/343

Hence, the ratio of their volumes is 27 : 343.

**(ii) Surface area of a sphere = 4πr**^{2}

^{2}

Ratio of surface areas of the spheres = 4πr_{1}^{2}/4πr_{2}^{2}

= r_{1}^{2}/r_{2}^{2}

= 3^{2}/7^{2}

= 9/49

Hence, the ratio of the surface areas is 9:49.

**(b) Given ratio of volume of two spheres = 125/64**

(4/3)πr_{1}^{3}/(4/3)πr_{2}^{3} = 125/64

⇒ r_{1}^{3}/ r_{2}^{3} = 125/64

Taking cube root on both sides

r_{1}/r_{2} = 5/4

Ratio of surface areas of the spheres = 4πr_{1}^{2}/4πr_{2}^{2}

= r_{1}^{2}/r_{2}^{2}

= 5^{2}/4^{2}

= 25/16

Hence, the ratio of the surface areas is 25:16.

**Question 10. ****Find the volume of a sphere whose surface area is 154 cm².**

**Answer :**

Given that

Surface area of a sphere = 154 cm²

4πr^{2} = 154

⇒ 4 × (22/7) × r^{2} = 154

⇒ r^{2} = (154 × 7)/(4 × 22)

⇒ r^{2} = 49/4

⇒ r = 7/2

Volume of the sphere = (4/3)πr^{3}

= (4/3) × (22/7) × (7/2)^{3}

= 539/3

= 179.666

= 179.67 cm^{3} (approx)

Hence the volume of the sphere is 179.67 cm^{3}

**Question 11. ****If the volume of a sphere is 179.2/3. Find its radius and surface area.**

**Answer :**

Given volume of the sphere is 179.2/3

(4/3)πr^{3} = 179.2/3 = 539/3

⇒ (4/3) × (22/7) × r^{3 }= 539/3

⇒ r^{3} = (539 × 3 × 7)/(4 × 22 × 3)

⇒ r^{3} = 49 × 7/8

⇒ r^{3} = (7 × 7 × 7)/(2 × 2 × 2)

Taking cube root on both sides, we get

r = 7/2 = 3.5 cm

Surface area of the sphere = 4r^{2}

**= **(4/3) × (22/7) × (7/2)^{2}

= 22 × 7

= 154 cm^{2}

Hence, the radius and the surface area of the sphere is 3.5 cm and 154 cm^{2} respectively.

**Question 12. ****A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?**

**Answer :**

Radius of a hemispherical bowl (r) = 3.5 cm

= 7/2 cm

Volume of the hemisphere = (2/3)πr^{3}

= (2/3) × (22/7) × (7/2)^{3}

= 11 × 49/6

= 539/6

= 89. 5/6 cm^{3}

Hence, the volume of the hemispherical bowl is 89.5/6 cm^{3}.

**(ML Aggarwal Mensuration Exe-17.3 Class 10)**

**Question 13. ****The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.**

**Answer :**

Given surface area of the sphere = 1256 cm^{2}

4πr^{2} = 1256

⇒ 4 × 3.14 × r^{2} = 1256

⇒ r^{2} = 1256 × /3.14 × 4

⇒ r^{2} = 100

⇒ r = 10 cm

Total surface area of the hemisphere = 3πr^{2}

= 3 × 3.14 × 10^{2}

= 3 × 3.14 × 100

= 942 cm^{2}

Hence,

the total surface area of the hemisphere is 942 cm^{2}.

Volume of the hemisphere = (2/3)πr^{3}

= (2/3) × 3.14 × 10^{3}

= (2/3) × 3.14 × 1000

= (2/3) × 3140

= 6280/3

= 2093.1/3 cm^{3}

Hence, the volume of the hemisphere is 2093.1/3 cm^{3}

-: End of ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions :–

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