13. Surface Areas and Volumes-Exercise 13.2 Mathematics Class 9 In English
Last Updated : 06 March 2026
Get updated solved mathematics for class 9 chapter 13. Surface Areas and Volumes Topic Exercise 13.2 with clear explanations free study material in English medium.
13. Surface Areas and Volumes-Exercise 13.2 Mathematics Class 9 In English
Last Updated : 06 March 2026
13. Surface Areas and Volumes-Exercise 13.2 Mathematics Class 9 In English
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)
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NCERT Solutions for Class 9 mathematics Chapter 13. Surface Areas and Volumes Topic Exercise 13.2
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13. Surface Areas and Volumes-Exercise 13.2 Mathematics Class 9 In English
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