NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4.

- Surface Areas and Volumes Class 9 Ex 13.1
- Surface Areas and Volumes Class 9 Ex 13.2
- Surface Areas and Volumes Class 9 Ex 13.3
- Surface Areas and Volumes Class 9 Ex 13.4
- Surface Areas and Volumes Class 9 Ex 13.5
- Surface Areas and Volumes Class 9 Ex 13.6
- Surface Areas and Volumes Class 9 Ex 13.7
- Surface Areas and Volumes Class 9 Ex 13.8
- Surface Areas and Volumes Class 9 Ex 13.9

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 13 |

Chapter Name |
Surface Areas and Volumes |

Exercise |
Ex 13.4 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

**Assume π = \(\cfrac { 22 }{ 7 } \) unless stated otherwise.**

Ex 13.4 Class 9 Maths Question 1.

Find the surface area of a sphere of radius :

(i) 5 cm (ii) 5.6 cm (iii) 14 cm

Solution.

^{
}

Ex 13.4 Class 9 Maths Question 2.

Find the surface area of a sphere of diameter:

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Solution.

Ex 13.4 Class 9 Maths Question 3.

Find the total surface area of a hemisphere of radius 10 cm (use π = &14)

Solution.

We have, radius of a hemisphere (r) = 10 cm

∴ Total surface area of a hemisphere = 3πr^{2
}= 3 x 3.14 x (10)^{2} = 9.42 x 100 = 942 cm^{2
}

Ex 13.4 Class 9 Maths Question 4.

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

Solution.

Let initial radius, r_{1} = 7 cm

After increases, r_{2} = 14 cm

Surface area for initial balloon = 4πr_{1}^{2} = 4x \(\cfrac { 22 }{ 7 } \) x 7 x 7 = 88 x 7

A_{1} = 616 cm^{2
}Surface area for increasing balloon 22

4πr_{2}^{2} = 4x \(\cfrac { 22 }{ 7 } \) x 14 x 14 = 88 x 28

A_{2} = 2464 cm^{2
∴} Required ratio = A_{1} : A_{2} = 616 :2464 = 1:4

Ex 13.4 Class 9 Maths Question 5.

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

Solution.

Ex 13.4 Class 9 Maths Question 6.

Find the radius of a sphere whose surface area is 154 cm2.

Solution.

Ex 13.4 Class 9 Maths Question 7.

The diameter of the moon Is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution.

Ex 13.4 Class 9 Maths Question 8.

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved, surface area of the bowl.

Solution.

The inner radius of the hemispherical bowl = 5 cm

Thickness of the hemispherical bowl

= 0.25 cm

cm

We know that,

Outer radius of the bowl

= Inner radius + Thickness

= (5+0.25) = & 25 cm

Outer curved surface area of the bowl = 2πr^{2}

= 2 x \(\cfrac { 22 }{ 7 } \) x 5.25 x 5.25 = 173.25 cm^{2}

Ex 13.4 Class 9 Maths Question 9.

A right circular cylinder just encloses a sphere of radius r (see figure). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).

Solution.

The radius of the sphere = r

Radius of the cylinder = Radius of the sphere = r

Height of the cylinder = Diameter = 2r

**(i)** Surface area of the sphere A_{1} = 4 πr^{2
}**(ii)** Curved surface area of the cylinder = 2πrh

A_{2} = 2π x r x 2r

A_{2} = 4πr^{2
}**(iii)** Required ratio = A_{1} : A_{2} = 4 π r^{2} : 4π r^{2} =1:1

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