#### Class nine NCERT solutions Surface Area and Volume

## 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 m**^{2}.

**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 m^{2}

Cost of painting 1 m^{2}
= 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 m**^{2}**. 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 m^{2},
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 m^{2}.

**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 m^{2}

=
22 × 5 m^{2} = 110 m^{2}

(ii)
Cost of plastering 1 m^{2} = ` 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) m^{2}

= 44x 0 025x
28x 025/ 7 m^{2} = 4.4 m^{2} (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 m^{2} = 59.4 m^{2}

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

= 2 × 22 7 × 2.1 (4.5 + 2.1) m^{2}

= 44 × 0.3 × 6.6 m^{2} = 87.12 m^{2}

Let the actual area of steel used be x m^{2}.

Area of steel wasted = 1/12
of x m^{2}
= x/12 m^{2} …. (i)

∴ Area of the steel used in the tank

Hence, 95.04 m^{2} of steel was actually used.