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.

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

**Question 1.**

**Find the surface area of a sphere of radius**

**(i) 10.5 cm**

**(ii) 5.6 cm**

**(iii) 14 cm**

**Solution:**

(i) We have, r = 105 cm

**Question 2.**

**Find the surface area of a sphere of diameter**

**(i) 14 cm**

**(ii) 21 cm**

**(iii) 3.5 m**

**Solution:**

**Question 3.**

**Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)**

**Solution:**

We have, 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}

**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} = 4 x \(\frac { 22 }{ 7 }\) x 7 x 7 = 88 x 7

A_{1} = 616 cm^{2}

Surface area for increasing balloon = 4πr_{2}^{2} = 4x \(\frac { 22 }{ 7 }\) x 14 x 14 = 88 x 28

A_{2} = 2464 cm^{2}

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

**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:**

We have, inner diameter = 10.5 cm

Inner radius = \(\frac { 10.5 }{ 2 }\) cm = 5.25 cm

Curved surface area of hemispherical bowl of inner side = 2πr

^{2}

**Question 6.**

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

**Solution:**

Surface area of a sphere = 154 cm

^{2}

Hence, the radius of the sphere is 3.5 cm.

**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:**

Let diameter of the Earth = d_{1}

**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:**

Outer radius of the bowl = (Inner radius + Thickness)

= ( 5 + 0.25) cm = 5.25 cm

**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} = 4πr^{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} = 4πr^{2 }: 4πr^{2} = 1 : 1

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