# NCERT Solutions for Exercise 13.2 Class 9 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths exercise 13.2 deals with the concept of the right circular cylinder and it’s surface areas . A right circular cylinder is a cylinder with a closed circular surface, two parallel bases on both ends, and elements that are perpendicular to the base. Three pieces make up the right circular cylinder. They are

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• Top circular base

• Curved lateral face

• Bottom circular face

In exercise 13.2 Class 9 Maths, A combination of two circles plus a rectangle forms the right circular cylinder. The surface area of a right circular cylinder is the area covered by the surface of the right circular cylinder. Surface area can be divided into two categories. They are

• Lateral Surface Area

• Total Surface Area

The curved surface area of the right circular cylinder, also known as the lateral surface area of the right circular cylinder, is the area covered by the curved surface of the cylinder. The total surface area of the right circular cylinder is the area occupied by the complete cylinder. NCERT solutions for Class 9 Maths chapter 13 exercise 13.2 consists of 11 questions which are direct formula based questions. The concepts related to the surface area and volumes are well explained in this NCERT syllabus Class 9 Maths chapter 13 exercise 13.2 . Along with NCERT book Class 9 Maths chapter 13 exercise 13.2 the following exercises are also present.

• Surface Area and Volumes Exercise 13.1
• Surface Area and Volumes Exercise 13.3
• Surface Area and Volumes Exercise 13.4
• Surface Area and Volumes Exercise 13.5
• Surface Area and Volumes Exercise 13.6
• Surface Area and Volumes Exercise 13.7
• Surface area and Volumes Exercise 13.8
• Surface Area and Volumes Exercise 13.9

## Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.2

Q1 The curved surface area of a right circular cylinder of height is . Find the diameter of the base of the cylinder.

Given,

The curved surface area of the cylinder =

And, the height of the cylinder,

We know, Curved surface area of a right circular cylinder =

Therefore, the diameter of the cylinder =

Q2 It is required to make a closed cylindrical tank of height and base diameter from a metal sheet. How many square metres of the sheet are required for the same?

Given,

Height of the cylindrical tank =

Base diameter =

We know,

The total surface area of a cylindrical tank =

Therefore, square metres of the sheet is

Q3 (i) A metal pipe is long. The inner diameter of a cross section is , the outer diameter being (see Fig. ). Find its inner curved surface area,

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Inner curved surface area =

Therefore, the inner curved surface area of the cylindrical pipe is

Q3 (ii) A metal pipe is long. The inner diameter of a cross section is , the outerdiameter being (see Fig. ). Find its outer surface area.

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Outer curved surface area =

Therefore, the outer curved surface area of the cylindrical pipe is

Q3 (iii) A metal pipe is long. The inner diameter of a cross section is 4 cm, the outer diameter being (see Fig. ). Find its total surface area.

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Outer curved surface area =

Inner curved surface area =

Area of the circular rings on top and bottom =

The total surface area of the pipe =

Therefore, the total surface area of the cylindrical pipe is

Q4 The diameter of a roller is and its length is . It takes complete revolutions to move once over to level a playground. Find the area of the playground in .

Given,

The diameter of the cylindrical roller =

Length of the cylindrical roller =

The curved surface area of the roller =

Area of the playground =

Therefore, the required area of the playground =

Q5 A cylindrical pillar is in diameter and in height. Find the cost of painting the curved surface of the pillar at the rate of per .

Given,

Radius of the cylindrical pillar, r =

Height of the cylinder, h =

We know,

Curved surface area of a cylinder =

Curved surface area of the pillar =

Now,

Cost of painting of the pillar =

Cost of painting the curved surface area of the pillar =

Therefore, the cost of painting curved surface area of the pillar is

Q6 Curved surface area of a right circular cylinder is . If the radius of the base of the cylinder is , find its height.

Given, a right circular cylinder

Curved surface area of the cylinder =

The radius of the base =

Let the height of the cylinder be

We know,

Curved surface area of a cylinder of radius and height =

Therefore, the required height of the cylinder is

Q7 (i) The inner diameter of a circular well is . It is deep. Find its inner curved surface area.

Given,

The inner diameter of the circular well =

Depth of the well =

We know,

The curved surface area of a cylinder =

The curved surface area of the well =

Therefore, the inner curved surface area of the circular well is

Q7 (ii) The inner diameter of a circular well is . It is deep. Find the cost of plastering this curved surface at the rate of Rs 40 per .

Given,

The inner diameter of the circular well =

Depth of the well =

The inner curved surface area of the circular well is

Now, the cost of plastering the curved surface per = Rs. 40

Cost of plastering the curved surface of =

Therefore, the cost of plastering the well is

Q8 In a hot water heating system, there is a cylindrical pipe of length and diameter . Find the total radiating surface in the system.

Given,

Length of the cylindrical pipe =

Diameter =

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius and length =

CSA of this pipe =

Therefore, the total radiating surface of the system is

Q9 (i) Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is in diameter and high.

Given, a closed cylindrical petrol tank.

The diameter of the tank =

Height of the tank =

We know,

The lateral surface area of a cylinder of radius and height =

The lateral surface area of a cylindrical tank =

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is

Q9 (ii) Find: how much steel was actually used, if of the steel actually used was wasted in making the tank.

Given, a closed cylindrical petrol tank.

The diameter of the tank =

Height of the tank =

Now, Total surface area of the tank =

Now, let of steel sheet be actually used in making the tank

Since of steel was wasted, the left of the total steel sheet was used to made the tank.
The total surface area of the tank =

Therefore, of steel was actually used in making the tank.

Q10 In Fig. , you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of and height of . A margin of 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.

Given, a cylindrical lampshade

The diameter of the base =

Height of the cylinder =

The total height of lampshade=

We know,

Curved surface area of a cylinder of radius and height =

Now, Cloth required for covering the lampshade = Curved surface area of the cylinder

Therefore, cloth will be required for covering the lampshade.

Q11 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 and height . 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?

Given, a cylinder with a base.

The radius of the cylinder =

Height of the cylinder =

We know,

The lateral surface area of a cylinder of radius and height =

Area of the cylindrical penholder = Lateral areal + Base area

Area of 35 penholders =

Therefore, the area of carboard required is

## More about NCERT Solutions for Class 9 Maths Exercise 13.2 :

The NCERT solutions for Class 9 Maths exercise 13.2 is mainly focused on the surface area of the right circular cylinder. In Exercise 13.2 Class 9 Maths, the lateral surface area of the closed right circular cylinder can be calculated by the product of 2π and the radius and height of the cylinder. CSA= 2πrh . The total surface area of a closed right circular cylinder is the sum of the area of the lateral surface and the area of the two bases.

TSA= CAS + 2(Area of a circle)

= 2πrh+2πr2

TSA=2πr(h+r)

The volume of the right circular cylinder was calculated by the product of the area of a circle and the height of the cylinder. V=πr2h. Surface areas are measured in square units, but the volume of a cube is measured in cubic units. In NCERT solutions for Class 9 Maths Exercise 13.2, the formulas for computing surface areas and volume for the correct circular cylinder are thoroughly covered.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.2 :

• NCERT solutions for Class 9 Maths exercise 13.2 gives a clear outline of the right circular cylinder in detail so that we get a clear understanding of what it is.

• By answering the NCERT solution for Class 9 Maths chapter 13 exercise 13.2 exercises, we can improve our grades in the first and second terms of school, as well as in our higher education.

• In exercise 13.2 Class 9 Maths, The formulas of calculating surface areas and volume for the right circular cylinder are well explained and these formulas will help in solving the competitive questions, hence the we must make a note of those formulas .

Also See:

• NCERT Solutions for Class 9 Maths Chapter 13 – Surface Area and Volumes

• NCERT Exemplar Solutions Class 9 Maths Chapter 13 – Surface Area and Volumes

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