NCERT solutions class 9 maths Chapter 13 Surface Areas and Volumes Exercise 13.2 are provided here. Our subject experts and teaching faculty together have prepared this NCERT Maths solution chapter wise, so that is beneficial for the students to solve the problems easily while using it as a reference. They also focus on creating solutions for these exercises in such a way that it is simple to understand for the students. Exercise 13.2 in Surface Areas and Volumes provides a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. NCERT solutions are prepared as per the NCERT guidelines so that it covers the entire syllabus from the examination point of view.

**Access other exercise solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes**

Exercise 13.1 solution (9 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.6 solution (8 questions)

Exercise 13.7 solution (9 questions)

Exercise 13.8 solution (10 questions)

Exercise 13.9 solution (3 questions)

**Access Answers of Maths NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2**

**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. (Assume π =22/7 )**

**Solution:**

Height of cylinder, h = 14cm

Let the diameter of the cylinder be d

Curved surface area of cylinder = 88 cm^{2}

We know that, formula to find Curved surface area of cylinder is 2πrh.

So 2πrh =88 cm^{2} (r is the radius of the base of the cylinder)

2×(22/7)×r×14 = 88 cm^{2}

2r = 2 cm

d =2 cm

Therefore, the diameter of the base of the cylinder is 2 cm.

**2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7**

**Solution:**

Let h be the height and r be the radius of a cylindrical tank.

Height of cylindrical tank, h = 1m

Radius = half of diameter = (140/2) cm = 70cm = 0.7m

Area of sheet required = Total surface are of tank = 2πr(r+h) unit square

= [2×(22/7)×0.7(0.7+1)]

= 7.48

Therefore, 7.48 square meters of the sheet are required.

**3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its**

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

**(ii) outer curved surface area**

**(iii) total surface area**

**(Assume π=22/7)**

**Solution:**

Let r_{1} and r_{2} Inner and outer radii of cylindrical pipe

r_{1 }= 4/2 cm = 2 cm

r_{2 }= 4.4/2 cm = 2.2 cm

Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) curved surface area of outer surface of pipe = 2πr_{1}h

= 2×(22/7)×2×77 cm^{2}

= 968 cm^{2}

(ii) curved surface area of outer surface of pipe = 2πr_{2}h

= 2×(22/7)×2.2×77 cm^{2}

= (22×22×2.2) cm^{2}

= 1064.8 cm^{2}

(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.

= 2r_{1}h+2r_{2}h+2π(r_{1}^{2}-r_{2}^{2})

= 9668+1064.8+2×(22/7)×(2.2^{2}-2^{2})

= 2031.8+5.28

= 2038.08 cm^{2}

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm^{2}.

**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}? (Assume π = 22/7)**

**Solution:**

A roller is shaped like a cylinder.

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

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2πrh

= 2×(22/7)×42×120

= 31680 cm^{2}

Area of field = 500×CSA of roller

= (500×31680) cm^{2}

= 15840000 cm^{2}

= 1584 m^{2}.

Therefore, area of playground is 1584 m^{2}. Answer!

**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 m ^{2}.**

**(Assume π = 22/7)**

**Solution:**

Let h be the height of a cylindrical pillar and r be the radius.

Given:

Height cylindrical pillar = h = 3.5 m

Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25cm = 0.25m

CSA of pillar = 2πrh

= 2×(22/7)×0.25×3.5

= 5.5 m^{2}

Cost of painting 1 m^{2} area = Rs. 12.50

Cost of painting 5.5 m^{2} area = Rs (5.5×12.50)

= Rs.68.75

Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m^{2} is Rs 68.75.

**6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)**

**Solution:**

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

Radius of the base of cylinder, r = 0.7m

CSA of cylinder = 2πrh

CSA of cylinder = 4.4m^{2}

Equating both the equations, we have

2×(22/7)×0.7×h = 4.4

Or h = 1

Therefore, the height of the cylinder is 1 m.

**7. The inner diameter of a circular well is 3.5m. It is 10m 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}.**

**(Assume π = 22/7)**

**Solution:**

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i) Inner curved surface area = 2πrh

= (2×(22/7 )×1.75×10)

= 110

Therefore, the inner curved surface area of the circular well is 110 m^{2}.

(ii)Cost of plastering 1 m^{2} area = Rs.40

Cost of plastering 110 m^{2} area = Rs (110×40)

= Rs.4400

Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

**8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find**

**the total radiating surface in the system. (Assume π = 22/7)**

**Solution:**

Height of cylindrical pipe = Length of cylindrical pipe = 28m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2×(22/7)×0.025×28 m^{2}

= 4.4m^{2}

The area of the radiating surface of the system is 4.4m^{2}.

**9. Find**

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

**diameter and 4.5m high.**

**(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)**

**Solution:**

Height of cylindrical tank, h = 4.5m

Radius of the circular end , r = (4.2/2)m = 2.1m

(i) the lateral or curved surface area of cylindrical tank is 2πrh

= 2×(22/7)×2.1×4.5 m^{2}

= (44×0.3×4.5) m^{2}

= 59.4 m^{2}

Therefore, CSA of tank is 59.4 m^{2}.

(ii) Total surface area of tank = 2πr(r+h)

= 2×(22/7)×2.1×(2.1+4.5)

= 44×0.3×6.6

= 87.12 m^{2}

Now, Let S m^{2} steel sheet be actually used in making the tank.

S(1 -1/12) = 87.12 m^{2}

This implies, S = 95.04 m^{2}

Therefore, 95.04m^{2 }steel was used in actual while making such a tank.

**10. In fig. 13.12, 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. (Assume π = 22/7)**

**Solution:**

Say h = height of the frame of lampshade, looks like cylindrical shape

r = radius

Total height is h = (2.5+30+2.5) cm = 35cm and

r = (20/2) cm = 10cm

Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh

= (2×(22/7)×10×35) cm^{2}

= 2200 cm^{2}

Hence, 2200 cm^{2} cloth is required for covering the lampshade.

**11. The students of 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? (Assume π =22/7)**

**Solution:**

Radius of the circular end of cylindrical penholder, r = 3cm

Height of penholder, h = 10.5cm

Surface area of a penholder = CSA of pen holder + Area of base of penholder

= 2πrh+πr^{2}

= 2×(22/7)×3×10.5+(22/7)×3^{2}= 1584/7

Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm^{2}

So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm^{2}

Therefore, 7920 cm^{2} cardboard sheet will be needed for the competition.

NCERT solutions for class 9 maths chapter 13 helps to find the surface area of various geometrical objects in a simplified and easy way. NCERT solutions for Class 9 Maths Exercise 13.2 explains how to find the surface area of the cylinder and helps to solve the problems based on it.

**Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2**

- These NCERT Solutions let you understand and solve all questions of Chapter 13.
- It helps to secure the best score in Maths second term exams.
- It contains all the important questions from the examination point of view.
- Good explanations help students to remember the concept in an effective way.