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

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 13 |

Exercise | Class 9 Chapter 13 Surface Areas and Volumes Exercise 13.4 |

Number of Questions Solved | 9 |

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

**NCERT TEXTBOOK EXERCISES**

**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 22/7 x 7 x 7 = 88 x 7

A_{1} = 616 cm^{2}

Surface area for increasing balloon = 4πr_{2}^{2} = 4x 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 = 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

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

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### Other Chapter of Class 9 Maths Chapter 13 Surface Areas and Volumes

**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**