NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.2 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

Exercise – 13.2

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. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Let the radius of the cylinder = r
Height of cylinder, h = 14 cm
Curved surface area of cylinder = 88 cm²
Curved Surface Area of a right circular cylinder = 2πrh
2πrh = 88 cm²
2 × $\frac{22}{7}$ × r × 14 cm = 88 cm²
r = $\frac{88\times 7}{2\times 22\times 14}$ = 1 cm

Diameter = 2 × radius
= 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 2 cm.

2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = $\mathbf{\frac{22}{7}}$

Answer – Height of the tank, h = 1 m
Diameter of cylindrical tank, d = 140 cm
Radius = half of diameter = (140/2) cm = 70cm = 0.7m
Total surface area of the closed cylindrical tank = 2πr(r + h)
= 2 × $\frac{22}{7}$ × 0.7 m × (0.7 m + 1 m)
= 4.4 m × 1.7 m
= 7.48 m²
7.48 m² of the sheet is required for making the cylindrical tank.

3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its

(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area
(Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Let r and R be the inner and outer radii of the cylindrical pipe
r= 4/2 cm = 2 cm
R = 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) Inner curved surface area = 2πrh
= 2 × $\frac{22}{7}$ × 2 × 77 cm²= 968 cm²

(ii) Outer curved surface area = 2πRh
= 2 × $\frac{22}{7}$ × 2.2 × 77 cm²= (22 × 22 × 2.2) cm²= 1064.8 cm²

(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
= 2πrh + 2πRh + 2π(R² – r²)
= 9668 + 1064.8 + 2 × $\frac{22}{7}$ × (2.2²– 2²) cm²
= 2031.8 + 2 × $\frac{22}{7}$ × 5.28 cm²= 2031.8 + 5.28 cm²
= 2038.08 cm²Therefore, the total surface area of the cylindrical pipe is 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? (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – A roller is shaped like a cylinder.
The Height of the roller, h = Length of roller = 120 cm
The Diameter of the roller, d = 84 cm
The Radius of the roller, r = Diameter/2 = 84/2 cm = 42 cm
Curved Surface Area of the roller = 2πrh
= 2 × $\frac{22}{7}$ × 42 cm × 120 cm
= 31680 cm²
Area of the playground = Area leveled by the cylinder in 500 revolutions
= 500 × 31680 cm²
= 15840000 cm²
= 15840000/10000 m² [Since 1cm² = 1/10000 m²]
= 1584 m²
Thus, area of the 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.
(Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
The Height of a cylindrical pillar, h = 3.5 m
The Diameter of a cylindrical pillar, d = 50 cm = 50/100 m = 0.50 m
The Radius of a cylindrical pillar, r = Diameter/2 = 0.50/2 =0.25 m
Curved surface area of the pillar = 2πrh
= 2 × $\frac{22}{7}$ × 0.25 m × 3.5 m
= 5.5 m²
Cost of painting the curved surface area at ₹12.50 per m^² = 12.50 × 5.5
= ₹ 68.75
Thus, cost of painting the curved surface of the pillar is ₹ 68.75.

6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – The curved surface area of the pillar = 4.4 m²
Radius of the cylinder, r = 0.7 m
Height of the cylinder = h
2πrh = 4.4 m²
2 × $\frac{22}{7}$ × 0.7 m × h = 4.4 m²
h = $\frac{7\times 4.4}{2\times22\times0.7}$
h = 1 m
The height of the right circular cylinder is 1 m.

7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m².
(Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
Inner Diameter of the circular well, d = 3.5 m
Inner Radius of the circular well, r = d/2 = 3.5/2 m = 1.75 m
Depth of circular well, say h = 10 m

(i) Inner curved surface area = 2πrh
= 2 × $\frac{22}{7}$ × 1.75 × 10
= 110 m²Therefore, the inner curved surface area of the circular well is 110 m².

(ii) Cost of plastering 1 m² area = Rs. 40
Cost of plastering 110 m² area = Rs (110 × 40)
= Rs. 4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
Diameter of the pipe, d = 5 cm
Radius of the pipe, r = d/2 = 5/2 cm = 2.5 cm = 2.5/100 m = 0.025 m
Length of the pipe, h = 28 m
The total radiating surface area of the pipe = 2πrh
= 2 × $\frac{22}{7}$ × 0.025 m × 28 m
= 4.4 m²
Thus, the total radiating surface is 4.4 m².

9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.
(ii) How much steel was actually used, if $\mathbf{\frac{1}{12}}$ of the steel actually used was wasted in making the tank. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
Diameter of the tank, d = 4.2 m
Radius of the tank, r = d/2 = 4.2/2 m = 2.1 m
Height of the tank, h = 4.5 m

(i) the lateral or curved surface area of the cylindrical tank = 2πrh
= 2 × $\frac{22}{7}$ × 2.1 × 4.5 m²= 44 × 0.3 × 4.5 m²= 59.4 m²

(ii) Total surface area of tank = 2πr(r+h)
= 2 × $\frac{22}{7}$ × 2.1 × (2.1 + 4.5)
= 44 × 0.3 × 6.6
= 87.12 m²Let the amount of steel required to make the tank be ‘x’.
Amount of steel required – Amount of steel wasted = Total surface area of the tank
x – $\frac{x}{12}$ = 87.12 m² [Since, 1/12 of the steel was wasted]
$\frac{11x}{12}$ = 87.12 m²
x = $\frac{12}{11}$ × 87.12 m²
x = 95.04 m²
Thus, Curve surface area = 59.4m², Steel actually used to make the tank = 95.04m²

10. 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. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Diameter of the frame, d = 20 cm
Radius of the frame, r = d/2 = 20/2 cm = 10 cm
The frame has a base height, h = 30 cm
Height of the lampshade, h = 30 cm + 2.5 cm + 2.5 cm = 35 cm
Thus, cloth required for covering the lampshade = 2πrh
= 2 × $\frac{22}{7}$ × 10 cm × 35 cm
= 2200 cm²
Thus, the area of the cloth required is 2200 cm².

11. The students of 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? (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
Radius of the penholder, r = 3 cm
Height of the penholder, h = 10.5 cm
Area of cardboard required for each penholder = 2πrh + πr² = πr(2h + r)
Area of cardboard required for 35 penholders = 35 × πr (2h + r )
= 35 × πr(2h + r )
= 35 × $\frac{22}{7}$ × 3 cm × (2 × 10.5 cm + 3 cm)
= 330 cm × 24 cm
= 7920 cm²
Therefore, 7920 cm² cardboard sheet will be needed for the competition.

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2