NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 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 13.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 13 |

Chapter Name |
Surface Areas and Volumes |

Exercise |
Ex 13.2 |

Number of Questions Solved |
11 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

**Question 1.**

**The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder.**

**Solution:**

We have, height = 14 cm

Curved surface area Of a right circular cylinder = 88 cm

^{2}

r = 1 cm

Diameter = 2 x Radius = 2 x 1 = 2 cm

**Question 2.**

**It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?**

**Solution:**

Let r be the radius and h be the height of the cylinder.

Given, r = \(\frac { 140 }{ 2 }\) = 70 cm = 0.70 m

and h = 1 m

Metal sheet required to make a closed cylindrical tank

= Total surface area

= 2πr (h + r)

= 2 x \(\frac { 22 }{ 7 }\) x 0.7(1 + 0.70)

= 2 x 22 x 0.1 x 170 = 7.48m^{2}

Hence, the sheet required to make a closed cylindrical tank = 7.48m^{2}

**Question 3.**

**A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its**

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

**(ii) outer curved surface area.**

**(iii) total surface area.**

**Solution:**

We have, h = 77cm

Outer diameter (d_{1}) = 4.4 cm

and inner diameter (d_{2}) = 4 cm

Outer radius (r_{1}) = 2.2 cm

Inner radius (r_{2}) = 2cm

(i) Inner curved surface area = 2πr^{2}h = 2x \(\frac { 22 }{ 7 }\) x 2 x 77 = 88 x 11 = 968 cm^{2}

(ii) Outer curved surface area = 2πr_{1}h

= 2 x \(\frac { 22 }{ 7 }\) x 2.2 x 77

= 44 x 2.2 x 11= 1064.8cm^{2}

(iii) Total surface area = Inner curved surface area + Outer curved surface area + Areas of two bases

= 968 + 1064.8 + 2\(\frac { 22 }{ 7 }\) (r_{1}^{2} – r^{2})

= 968 + 1064.8 + 2 x \(\frac { 22 }{ 7 }\) [(2.2) – r^{2}]

= [2032.8 + 2 x \(\frac { 22 }{ 7 }\) (4.84 – 4)]

= 2032.8 +\(\frac { 44 }{ 7 }\) x 0.84 = 2032.8+ 44x 0.12

= 2032.8 + 5.28 cm^{2} = 2038.08 cm^{2}

**Question 4.**

**The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m ^{2}.**

**Solution:**

We have, diameter of a roller = 84 cm

r = radius of a roller = 42 cm

h = 120 cm

To cover 1 revolution = Curved surface area of roller

= 2πrh

= 2 x \(\frac { 22 }{ 7 }\) x 42 x 120

= 44 x 720 cm

^{2}

= 31680 cm

^{2}

Area of the playground = Takes 500 complete revolutions = 500 x 3.168 m

^{2}

= 1584 m

^{2}

**Question 5.**

**A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m ^{2}.**

**Solution:**

Given, Diameter = 50 cm

Curved surface area of the pillar = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 0.25 x 35

= 2 x 22 x 0.25 x 0.5 = 55 m

^{2}

Cost of painting per m2 = ₹ 12.50

Cost of painting 5.5 m2 = ₹ 12.50×5.5 = ₹ 68.75

**Question 6.**

**Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.**

**Solution:**

We have, curved surface area of a right circular cylinder = 4.4m

^{2}

∴ 2πrh = 4.4

Hence, the height of the right circular cylinder is 1 m

**Question 7.**

**he inner diameter of a circular well is 3.5 m. It is 10 m deep. Find**

**(i) its inner curved surface area.**

**(ii) the cost of plastering this curved surface at the rate of ₹40 per m ^{2}.**

**Solution:**

We have, inner diameter = 3.5 m

∴ inner radius = \(\frac { 3.5 }{ 2 }\) m

and h= 10m

(i) Inner curved surface area = 2πrh = 2x \(\frac { 22 }{ 7 }\) x \(\frac { 3.5 }{ 2 }\) x10 = 22 x 5 = 110m

^{2}

(ii) Cost of plastering perm2 = ₹40

Cost of plastering 110 m2 = ₹40x 110= ₹4400

**Question 8.**

**In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.**

**Solution:**

We have, h = 28 m

Diameter = 5 cm

Total radiating surface in the system = Curved surface area of the cylindrical pipe

= 2πrh =2 x \(\frac { 22 }{ 7 }\) x 0.025 x 28 = 4.4 m^{2}

**Question 9.**

**Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.**

**(ii) how much steel was actually used, if \(\frac { 1 }{ 12 }\) of the steel actually used was wasted in making the tank?**

**Solution:**

(i) We have, diameter = 4.2 m

Since, \(\frac { 1 }{ 2 }\) of the actual steel used was wasted, therefore the area of the steel which was actually used for making the tank

**Question 10.**

**In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.**

**Solution:**

Given r = \(\frac { 20 }{ 2 }\) cm = 10cm

h = 30 cm

Since, a margin of 2.5 cm is used for folding it over the top and bottom so the total height of frame,

h1 = 30 + 25 + 25

h1 = 35 cm

∴ Cloth required for covering the lampshade = Its curved surface area

**Question 11.**

**The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?**

**Solution:**

Cardboard required by each competitor

= Base area + Curved surface area of one penholder

= πr2 + 2πrh [Given, h = 10.5 cm, r = 13cm]

= \(\frac { 22 }{ 7 }\) x (3)2 + 2 x \(\frac { 22 }{ 7 }\) x 3 x 105

= (28.28 + 198) cm^{2}

= 22828 cm^{2}

For 35 competitors cardboard required = 35 x 22628 = 7920 cm^{2}

Hence, 7920 cm^{2} of cardboard was required to be bought for the competition.

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