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## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.2

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.

Sol. Here, r = 50/2 cm = 25 cm = 0.25 m, h = 3.5 m

Curved surface area of the pillar

= 2πrh = 2 × 22/7 × 0.25 × 3.5 m2 = 5.5 m2

Cost of painting 1 m2 = Rs 12.50

∴ Total cost of painting the curved surface of the pillar = 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 m, find its height.

Sol. The curved surface area of the cylinder

= 4.4 m2, r = 0.7 m, h = ?

The curved surface area of the cylinder = 2πrh

⇒ 4.4 = 2 × 22/7 × 0.7 × h

⇒ h = 4.4/4.4  = 1

Hence, the height of the cylinder is 1 m

7. 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.

Sol. Here, r = 3.5/2 m , h = 10 m

(i) The inner curved surface area of the well

= 2πrh

= 2 × 22/7 × 3.5/ 2. × 10 m2

= 22 × 5 m2 = 110 m2

(ii) Cost of plastering 1 m2 = ` 40

∴ Cost of plastering the curved surface area of the well = Rs 110 × 40 = 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.

Sol. Here, r = 52 cm = 2.5 cm = 0.025 m, h = 28 m.

Total radiating surface in the system

= total surface area of the cylinder

= 2π r(h + r) = 2× 22/7 × 0.025 (28 + 0.025) m2

= 44x 0 025x 28x 025/ 7 m2 = 4.4 m2 (approx)

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 1/12 of the steel actually used was wasted in making the tank?

Sol. Here, r = 4.2/2 m = 2.1 m, h = 4.5 m

(i) Curved surface area of the storage tank

= 2πrh = 2 × 22 7 × 2.1 × 4.5 m2 = 59.4 m2

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

= 2 × 22 7 × 2.1 (4.5 + 2.1) m2

= 44 × 0.3 × 6.6 m2 = 87.12 m2

Let the actual area of steel used be x m2.

Area of steel wasted = 1/12 of x m2 = x/12 m2 …. (i)

∴ Area of the steel used in the tank

Hence, 95.04 m2 of steel was actually used.

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