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 cm^{2}. Find the diameter of the base of the cylinder.

Answer:

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 cm^{2}

⇒ ⇒ 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?

Answer:

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) m^{2}

= (2 × 22 × 0.1 × 1.7) m^{2} = 7.48 m^{2}

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.

Answer:

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 cm^{2}

= 2 × \(\frac{22}{7}\) × 2 × 77 cm^{2}

= 968 cm^{2}

(ii) Outer curved surface

= 2πRh cm^{2}

= 2 × \(\frac{22}{7}\) × 2.2 × 77 cm^{2}

= 1064.8 cm^{2}

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π(R^{2} – r^{2})

= [968 + 1064.8 + (2 × \(\frac{22}{7}\)) (4.84 – 4)] cm^{2}

= (2032.8 + \(\frac{44}{7}\) × 0.84 7)

= (2032.8 + 5.28) cm^{2}

= 2038.08 cm^{2}

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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.

Answer:

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) m^{2}

= 3.1680 m^{2}

Area of the playground

= (500 × 3.168) m^{2} = 1584 m^{2}

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 m^{2}.

Ans. Radius of the pillar (r) = \(\frac{50}{2}\) = 25 cm = 0.25 m^{2}

Height of the pillar (h) = 3.5 m.

Rate of painting = ₹ 12.50 per m^{2}

Curved surface area

= 2πrh = (2 × \(\frac{22}{7}\) × 0.25 × 3.5) m^{2}

= 5.5 cm^{2}

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

Question 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.

Answer:

Let r be the radius of the base and h be the height of the cylinder.

Curved surface area = 2πrh = 4.4 m^{2}

⇒ 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 m^{2}.

Answer:

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 m^{2}

(i) Curved surface area = 2πrh

= (2 × \(\frac{22}{7}\) × 1.75 × 10) m^{2}

= 110 m^{2}

(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.

Answer:

Radius of the pipe (r) = \(\frac{5}{2}\) cm = 2.5 cm = 0.025 m^{2}

Length of cylindrical pipe (h) = 28 m

Total radiating surface = Curved surface area of the pipe

= 2πrh

= (2 × \(\frac{22}{7}\) × 0.025 × 28) m^{2}

= 4.4 m^{2}

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

Answer:

(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 m^{2} = 2 × \(\frac{22}{7}\) × 2.1 × 4.5 7

= 59.4 m^{2}

(ii) Total surface area of the tank = 2πr(r + h) m^{2}

= [2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5)] m^{2}

= 87.12 m^{2}

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 m^{2}

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.

Answer:

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) cm^{2} = 2200 cm^{2}

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?

Answer:

Radius of the penholder (r) = 3 cm

Height of the penholder (h) = 10.5 cm

Cardboard required by 1 competitor = CSA of one penholder + area of the base

= 2πrh + πr^{2}

= [(2 × \(\frac{22}{7}\) × 3 × 10.5) + \(\frac{22}{7}\) × 3^{2}] cm^{2}

= 198 + \(\frac{198}{7}\)

= \(\frac{1584}{7}\) cm^{2}

Cardboard required for 35 competitors = (35 × \(\frac{1584}{7}\)) cm^{2} = 7920 cm^{2}