Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Questions and Answers.

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

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.

Solution:

Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then,

Curved surface area = 2πrh

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?

Solution:

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

Metal sheet required to make a closed cylindrical tank = Its total surface area

Hence, the sheet required = 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 figure). Find its

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

Solution:

We have, R = external radius = \(\frac { 4.4 }{ 2 }\) cm = 2.2 cm

r = internal radius = \(\frac { 4 }{ 2 }\) cm = 2 cm

h = length of the pipe = 77 cm.

(i) Inner curved surface = 2πrh cm²

= 2 x \(\frac { 22 }{ 7 }\) x 2 x 77 cm²

= 968 cm²

(ii) Outer curved surface = 2πRh cm²

= 2 x \(\frac { 22 }{ 7 }\) x 2.2 x 77 cm²

= 1064.8 cm²

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

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 m².

Solution:

The length of the roller is 120 cm i.e., h = 1.2 m and radius of the cylinder (i.e., roller) = \(\frac { 84 }{ 2 }\) cm = 42 cm = 0.42 m.

Distance covered by roller in one revolution

= Its curved surface area

= 2πrh

= (2 x \(\frac { 22 }{ 7 }\) x 0.42 x 1.2) m²

= 3.168 m²

Area of the playground = Distance covered by roller in 500 revolution

= (500 x 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 Rs 12.50 per m².

Solution:

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

∴ r = \(\frac { 50 }{ 2 }\) cm = 25 cm = 25 m and h = 3.5 m.

Curved surface = 2πrh

= (2 x \(\frac { 22 }{ 7 }\) x 0.25 x 3.5) m²

= 5.5 m²

Cost of painting the curved surface @ Rs 12.50 per m²

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.

Solution:

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

Curved surface area = 4.4 m²

⇒ 2πrh = 4.4

⇒ 2 x \(\frac { 22 }{ 7 }\) x 0.7 x h = 4.4 [∵ r = 0.7 m]

⇒ h = \(\left(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\right)\) m = 1 m

Thus, the height of the cylinder = 1 metre.

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 Rs 40 per m².

Solution:

(i) Let r be the radius of the face and h be depth of the well. Then,

Curved surface = 2nrh

= (2 x \(\frac { 22 }{ 7 }\) x \(\frac { 3.5 }{ 2 }\) x 10) m² = 110 m²

(ii) Cost of plastering is Rs 40 per m²

∴Cost of plastering the curved surface = Rs (110 x 40)

= Rs 4400.

Question 8.

In a hot water heating system, there is a cylindrical pipe Pf length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Total radiating surface in the system

= Curved surface area of the pipe

= 2πrh,

where r = \(\frac { 5 }{ 2 }\) cm = 2.5 cm = \(\frac { 2.5 }{ 100 }\) m = 0.025 m and h = 28 m

= (2 x \(\frac { 22 }{ 7 }\) x 0.025 x 28) m² = 4.4 m²

Question 9.

Find (i) the lateral or curved surface.area of a 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 closed tank.

Solution:

(i) Here, r = (\(\frac { 4.2 }{ 2 }\))m = 2.1 m and h = 4.5 m.

(ii) Since \(\frac { 1 }{ 12 }\) of the actual steel used was wasted, the area of the steel which has gone into the tank = (1 – \(\frac { 1 }{ 12 }\)) of x = \(\frac { 11 }{ 12 }\) of x.

Hence, the actual area of the steel used = 95.04 m²

Question 10.

In figure, 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.

Solution:

Here r = (\(\frac { 20 }{ 2 }\)) cm = 10 cm and h = 30 cm + 2 x 2.5 cm (i.e margin) = 35 cm.

Cloth required for covering the lampshade

= Its curved surface area

= 2πrh

= (2 x \(\frac { 22 }{ 7 }\) x 10 x 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?

Solution:

Cardboard required by each competitor = Curved surface area of one penholder + base area

= 2πrh + πr² where r = 3 cm, h = 10.5 cm

= [(2 x \(\frac { 22 }{ 7 }\) x 3 x 10.5) + \(\frac { 22 }{ 7 }\) x 9] cm²

= (198 + 28.28) cm²

= 226.28 cm² (approx.)

Cardboard required for 35 competitors = (35 x 226.28) cm²

= 7920 cm² (approx.)