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

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

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Question 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.
Let r be the radius of the base and
h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm2
⇒ ⇒ 2 × $$\frac{22}{7}$$ × r × 14 = 88
⇒ r = $$\frac{88}{\left(2 \times \frac{22}{7} \times 14\right)}$$ = 1 cm
⇒ r = 1 cm
Thus, the diameter of the base = 2π = 2 ×1 = 2 cm

Question 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?
Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1 m

Radius of base (r) = $$\frac{140}{2}$$ = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × $$\frac{22}{7}$$ × 0.7) (1 + 0.7) m2
= (2 × 22 × 0.1 × 1.7) m2 = 7.48 m2

Question 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 Fig. ).

Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Let R be external radius and r be the internal radius and h be the length of the pipe.
R = $$\frac{4.4}{2}$$ = 2.2 cm
r = $$\frac{4}{2}$$ = 2 cm
h = 77 cm

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

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

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh + 2πRh + 2π(R2 – r2)
= [968 + 1064.8 + (2 × $$\frac{22}{7}$$) (4.84 – 4)] cm2
= (2032.8 + $$\frac{44}{7}$$ × 0.84 7)
= (2032.8 + 5.28) cm2
= 2038.08 cm2

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Question 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 nr.
Length of the roller (h) = 120 cm = 1.2 m
Radius of the roller (r) = $$\frac{88}{2}$$ cm
= 42 cm = 0.42 m

Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × $$\frac{22}{7}$$ × 42 × 1.2) m2
= 3.1680 m2

Area of the playground
= (500 × 3.168) m2 = 1584 m2

Question 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 ₹ 12.50 per m2.
Ans. Radius of the pillar (r) = $$\frac{50}{2}$$ = 25 cm = 0.25 m2
Height of the pillar (h) = 3.5 m.
Rate of painting = ₹ 12.50 per m2
Curved surface area
= 2πrh = (2 × $$\frac{22}{7}$$ × 0.25 × 3.5) m2
= 5.5 cm2

Total cost of painting = (5.5 × 12.5) = ₹ 68.75

Question 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.
Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m2
⇒ 2 × $$\frac{22}{7}$$ × 0.7 × h = 4.4
⇒ h = $$\frac{4.4}{\left(2 \times \frac{22}{7} \times 0.7\right)}$$ = 1 m

Question 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.
Radius of circular well (r) = $$\frac{3.5}{2}$$ m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = ₹ 40 per m2

(i) Curved surface area = 2πrh
= (2 × $$\frac{22}{7}$$ × 1.75 × 10) m2
= 110 m2

(ii) Cost of plastering = ₹ (110 × 40) = ₹ 4400

Question 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.
Radius of the pipe (r) = $$\frac{5}{2}$$ cm = 2.5 cm = 0.025 m2
Length of cylindrical pipe (h) = 28 m
Total radiating surface = Curved surface area of the pipe
= 2πrh
= (2 × $$\frac{22}{7}$$ × 0.025 × 28) m2
= 4.4 m2

Question 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 $$\frac{1}{12}$$ of the steel actually used was wasted in making the tank
(i) Radius of the tank (r) = $$\frac{4.2}{2}$$ m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m2 = 2 × $$\frac{22}{7}$$ × 2.1 × 4.5 7
= 59.4 m2

(ii) Total surface area of the tank = 2πr(r + h) m2
= [2 × $$\frac{22}{7}$$ × 2.1 (2.1 + 4.5)] m2
= 87.12 m2

Let x be the actual steel used in making tank.
∴ (1 – $$\frac{1}{12}$$) × x = 87.12
⇒ x = 87.12 × $$\frac{12}{11}$$ = 95.04 m2

Question 10.
In Fig., 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.

Radius of the frame (r) = $$\frac{20}{2}$$ cm = 10 cm

Height of the frame (h) = 30 cm + 2 × 2.5 cm = 35 cm
(2.5 cm of margin will be added both sides in the height.)

Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × $$\frac{22}{7}$$ × 10 × 35) cm2 = 2200 cm2

Question 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?
= [(2 × $$\frac{22}{7}$$ × 3 × 10.5) + $$\frac{22}{7}$$ × 32] cm2
= 198 + $$\frac{198}{7}$$
= $$\frac{1584}{7}$$ cm2
Cardboard required for 35 competitors = (35 × $$\frac{1584}{7}$$) cm2 = 7920 cm2