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

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

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

Ex 13.2 Class 9 Maths Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution.
Let radius of the base be r cm.
Given, height (h) = 14 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.1

Ex 13.2 Class 9 Maths 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 meters of the sheet are required for the same?
Solution.
Let r be the radius and h be the height of the cylinder.
Given, diameter of the base = 140 cm
∴ Radius (r) = \cfrac { 140 }{ 2 } y = 70 cm = 0.70 m and h = 1 m
Metal sheet required to make a closed cylinder tank
= Total surface area of right circular cylinder = 2πr(h + r)
= 2 x \cfrac { 22 }{ 7 } x 0.7 (1 + 0.7) = 2 x 22 x 0.1 x 1.7 = 7.48 m2
Hence, the sheet required to make a closed cylindrical tank is 7.48 m2.

Ex 13.2 Class 9 Maths Question 3.
A metal pipe is 77 cm long. The inner 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.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.2
Given, height of pipe (h) = 77 cm
Outer diameter (d1) = 4.4 cm
and inner diameter (d2) = 4 cm
∴ Outer radius (r1) = 2.2 cm  [∵  radius = \cfrac { diameter }{ 2 } ]
(i) Inner radius (r2) = 2 cm Inner curved surface area = 2πrh
[∵ curved surface area of a right circular cylinder = 2πrh]
= 2 x \cfrac { 22 }{ 7 } x 2 x 77 = 88 x 11 = 968 cm2

(ii) Outer curved surface area = 2πr,h
= 2 x \cfrac { 22 }{ 7 } x 2.2 x 77 = 44 x 2.2 x 11 = 1064.8 cm2

(iii) Total surface area = Inner curved surface area + Outer curved surface area + Area of two bases
= 968 +1064.8 + 2π(r12 – r22)
= 968+1064.8+ 2 x \cfrac { 22 }{ 7 } [[(2.2)2 -22]
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.3

Ex 13.2 Class 9 Maths 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 m2.
Solution.
We have, diameter of a roller = 84 cm
r = radius of a roller = 42 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.4

Ex 13.2 Class 9 Maths 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 Rs. 12.50 per m2.
Solution.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.5

Ex 13.2 Class 9 Maths Question 6.
Curved surface area of a right circular cylinder is 44 m2. 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 = 4.4 m2
2πrh = 4.4
⇒ 2 x \cfrac { 22 }{ 7 } x 0.7 x h = 4.4     (r=0.7m, given)
⇒ h = \cfrac { 44 }{ 44 }
⇒ h = 1m
Hence, the height of the right circular cylinder is 1 m.

Ex 13.2 Class 9 Maths Question 7.
The 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 Rs. 40 per m 2.
Solution.
We have, inner diameter = 3.5 m
∴ Inner radius = \cfrac { 3.5 }{ 2 }   m and height (h) = 10 m 2

(i) Inner curved surface area = 2πrh
= 2 x \cfrac { 22 }{ 7 } x \cfrac { 3.5 }{ 2 } x 10 = 22 x 5 = 110 m2

(ii) Given, cost of plastering per m2 = Rs. 40
∴ Cost of plastering 110 m2 of circular well = 40 x 110 = Rs. 4400.

Ex 13.2 Class 9 Maths 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.
Given, length or height of the pipe (h) = 28 m and diameter = 5 cm
Radius (r) = \cfrac { 5 }{ 2 } = 2.5 cm = \cfrac { 2.5 }{ 100 } m = 0.025 m            [∵  1 cm =\cfrac { 1 }{ 1000 } m]
Total radiating surface in the system
= Curved surface area of the cylindrical pipe 22
= 2πrh = 2 x \cfrac { 22 }{ 7 } x 0.025 x 28 = 4.4 m2

Ex 13.2 Class 9 Maths 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 \cfrac { 1 }{ 12 } of the steel actually used was wasted in making the tank?
Solution.
(i) We have, diameter = 4.2 m
Radius, r = \cfrac { 4.2 }{ 2 } = 2.1 m  and h = 4.5 m
Curved surface area of a closed cylindrical petrol storage tank
= 2πrh
= 2 x 2 x \cfrac { 22 }{ 7 } x 2.1 x 4.5 7
= 44 x 0.3 x 4.5
= 59.4 m2

(ii) Now, total surface area = Curved surface area + 2πr2
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.6

Ex 13.2 Class 9 Maths Question 10.
In given 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.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.7
Given, diameter of the base = 20 cm 20
∴ Radius (r) = \cfrac { 20 }{ 2 } = 10 cm and height (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 +2.5+ 2.5 = 35 cm
∴ Cloth required for covering the lampshade
= Curved surface area of lampshade = 2πr(h1)
= 2 x \cfrac { 22 }{ 7 } x 10 x (35) = \cfrac { 440 }{ 7 } x 35 = 440 x 5 = 2200 cm2

Ex 13.2 Class 9 Maths 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
= nπ2 +2πrh  [Given, h = 10.5 cm, r = 3 cm]
\cfrac { 22 }{ 2 } \times { (3) }^{ 2 }+2\times \cfrac { 22 }{ 7 } \times 3\times 10.5
= (28.28+ 198) cm2
= 226.28 cm2
For 35 competitors cardboard required = 35 x 226.28 = 7920 cm2
Hence, 7920 cm2 of cardboard was required to be bought for the competition.

 

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