# Surface Areas and Volumes – Exercise 13.2 – (MATHEMATICS) – 9th Class

**Exercise 13.2**

Assume x = unless stated otherwise.

**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.**

### Show Answer

**Sol.** Given: 2πrh =

⇒ r = 1 cm

∴ diameter of the base = 2 cm.

**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 metres of the sheet are required for the same?**

### Show Answer

**Sol.** Base diameter = 140 cm = 1.4 m. Height (*h*) = 1 m.

∴ Base radius (r) = 0.7 m

Total surface area (required metal sheet)

= 4.4 × 1.7 m^{2} = 7.48 m^{2}.

**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.**

### Show Answer

**Sol.** Length of the pipe (h) = 77 cm

Inner diameter = 4 cm, inner radius (r_{1}) = 2 cm, length = 77 cm,

outer diameter = 4.4 cm, outer radius (r_{2}) = 2.2 cm

(i) Inner curved surface area = 2πr_{1}h

(ii) Outer curved surface area = 2πr_{2}h

= 1064.8 cm².

(iii) Total surface area = 2π_{1}h + 2πr_{2}h + 2π(r_{2}^{2} – r_{1}^{2})

= 968 + 1064.8 + 2 ×

= 968 + 1064.8 + 5.28

= 2038.08 cm².

**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.**

### Show Answer

**Sol.** Area levelled in 1 revolution = 2πrh.

= 31680 cm².

in 500 revolutions (area of playground)

= 1584 m².

**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.**

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**Sol.** Curved surface of pillar =

= 5.5 m^{2}

Cost of painting = Rs.(12.50 × 5.5) = Rs.68.75.

**6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 in, find its height.**

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**Sol.**

**7. The inner diameter of a circular well is 3.5 m. It is 10 in deep. Find**

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

**(ii) the cost of plastering this curved surface at the rate of T40 per m ^{2}.**

### Show Answer

**Sol.** (i) Inner curved surface area = 2πrh

= 110 m^{2}.

(ii) Cost of plastering = Rs.(40 × 110) = Rs.4400.

**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.**

### Show Answer

**Sol.** Total radiating surface =

= 4.4 m^{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 4.5 m high.**

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

### Show Answer

**Sol.** (i) Lateral surface area =

(ii) Total surface area of the tank = 59.4 + 2 × × 2.1 × 2.1

= 59.4 + 27.72

= 87.12 m^{2}

Let total steel used = x m^{2 }

∴

= 95.04 m^{2}

**10. In the adjoining figure, you see the frame of a lampshade. .It is to he 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.**

### Show Answer

**Sol.** Total height including margin = (30 + 2 × 2.5) cm = 35 cm.

Since diameter = 20 cm therefore, radius = 10 cm.

Total cloth required

= 2200 cm².

**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?**

### Show Answer

**Sol.** Total cardboard required

= 35 (198 + 28.29) cm² = 7920 cm².