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13. Surface Areas and Volumes MATHEMATICS class 9 in English Medium ncert book solutions Exercise 13.2
13. Surface Areas and Volumes Exercise 13.2 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 13. Surface Areas and Volumes Exercise 13.2 to help you master concepts and score higher.
13. Surface Areas and Volumes MATHEMATICS class 9 in English Medium ncert book solutions Exercise 13.2
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 13. Surface Areas and Volumes clearly. This chapter includes the topic Exercise 13.2, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 13. Surface Areas and Volumes is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 13.2. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
13. Surface Areas and Volumes
Exercise 13.2
EXERCISE 13.2
Assume that π = 
, unless stated otherwise.
Q1. 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:
H = 14cm, curved surface area of cylinder = 88cm2
Curved surface area of cylinder = 88cm2
⇒ 2 × × r × h = 88cm2
⇒ 2 × × r × 14 = 88cm2
⇒ 88r = 88
⇒ r = 88/88
⇒ r = 1cm
D = 2r ⇒ 2 ×1 = 2cm
Q2. 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:
D = 140cm ⇒ r = 70cm = 0.7m, H = 1m
Total surface area of cylinder = 2πr(r + h)
⇒ 2 × × 0.7 (0.7 + 1)
⇒ 2 × 22 × 0.1 × 1.7
⇒ 7.48m2
Hence, 7.48m2 metal sheet required to make closed cylindrical tank.
Q3. 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 Fig. 13.11). Find its
(i) Inner curved surface area, 
(ii) Outer curved surface area,
(iii) Total surface area.
Solution:
H = 77cm, H = 77cm
D = 4cm, D = 4.4cm
R = 2cm, R = 2.2cm
(i) interior curved surface area = 2πrh
⇒ 2 × × 2 ×77
⇒ 88 × 11
⇒ 968 cm2
(ii) exterior curved surface area = 2πrh
⇒ 5.28 cm2
Total surface area = interior surface area + exterior surface area + ring surface area
⇒ 1064.8 + 9.68 + 5.28 cm2
⇒ 2038.08 cm2
Q4. 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:
D = 84cm ⇒ r = 42cm, H = 120cm
Curved surface area of cylinder = 2πrh
⇒ 2 × × 42 × 120
⇒ 44 × 6 × 120
⇒ 31680 cm2
Total area of playground
Q5. 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 m2.
Solution:
D = 50 cm , ⇒ R = 0.25m, H = 3.5m
Curved surface area of cylinder = 2πrh
Q6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Curved surface area of cylinder = 4.4 m2 , radius = 0.7 m
Let the height of the cylinder be = h
2πrh = 4.4
Q7. 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 40 per m2.
Solution:
Inner diameter of circular well = 3.5 m ⇒ r = 1.75 m
Depth of the well = 10 m
(i) inner surface area of the well = 2πrh
Q8. 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:
The length of the pipe = 28 m
Q9. 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 of the steel actually used was wasted in making the tank.
Solution:
(i) diameter of the cylindrical petrol tank = 4.2 m
Radius of the tank = 2.1m , height = 4.5 m
Curved surface area of cylinder = 2πrh
Q10. 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.
Solution:
Height of the folding = 2.5 cm
Height of the frame = 30 cm
Diameter = 20 cm ⇒ radius = 10 cm
Now cloth required for covering for lampshade
= C.S.A of top + C.S.A of middle + C.S.A of bottom
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 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:
Radius of pen holder = 3 cm
Height of pen holder = 10.5 cm
Cardboard required for pen holder = CSA of pen holder + area of circular base
⇒ 2πrh + πr2
⇒ πr(2h + r)
See other sub-topics of this chapter:
1. Exercise 13.1 2. Exercise 13.2 3. Exercise 13.3 4. Exercise 13.4 5. Exercise 13.5 6. Exercise 13.6 7. Exercise 13.7 8. Exercise 13.8 9. Exercise 13.9