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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

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

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

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.
Answer:
Height of cylinder h = 14 cm.
Curved surface area of a cylinder = 2 πrh
∴ 88 cm² = 2 × r × 14cm
∴ \(\frac{88 \times 7}{2 \times 22 \times 14}\) cm = r
∴ r = 1 cm
Now, diameter of the cylinder = 2r = 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 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:
Height of cylindrical tank h = 1 m
Diameter of the cylinder =140 cm
∴ Radius of the cylinder r = \(\frac{\text { diameter }}{2}\)
= \(\frac{140}{2}\) cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.7 (0.7 + 1) m²
= 4.4 × 1.7 m²
= 7.48 m²
Thus, 7.48 m² sheet is required to make the closed cylindrical tank.

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 the given figure). Find its

(i) inner curved surface area,
Answer:
For inner cylinder, diameter = 4 cm
∴ For inner cylinder,
radius r = \(\frac{\text { diameter }}{2}\) = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 968 cm²
Thus, the inner curved surface area is 968 cm².

(ii) outer curved surface, area,
Answer:
For outer cylinder, diameter = 4.4 cm
∴ For outer cylinder,
radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{4.4}{2}\) = 2.2
and height h = 77 cm.
Outer curved surface area of the pipe
= 2πRh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 1064.8 cm²
Thus, the outer curved surface area is
1064.8 cm².

(iii) total surface area.
Answer:
Total surface area includes the area of two circular rings at the ends together with the inner and outer curved surface areas.
For each circular ring, outer radius R = 2.2 cm and inner radius r = 2 cm
Area of one circular ring
= π(R² – r²)
= \(\frac{22}{7}\)(2.2² – 2²)cm²
= \(\frac{22}{7}\) (4.84 – 4) cm²
= \(\frac{22}{7}\) × 0.84 cm²
= 2.64 cm²
∴ Area of two circular rings.
= 2 × 2.64 cm²
= 5.28 cm²
Now, total surface area of the pipe = Inner curved surface area + outer curved surface area + area of two circular rings
= 968 + 1064.8 + 5.28 cm²
= 2038.08 cm²

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.
Answer:
For the cylindrical roller, diameter d = 84 cm and height (length) h = 120 cm.
Curved surface area of the cylindrical roller
= πdh
= \(\frac{22}{7}\) × 84 × 120 cm²
= 31680 cm²
= \(\frac{31680}{10000}\) m²
= 3.168 m²
Thus, the area of playground levelled in 1 complete revolution of the roller = 3.168 m²
∴ The area of playground levelled in 500 complete revolutions of the roller
= 3.168 × 500 m² = 1584 m²
Thus, the area of the playground is 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.
Answer:
For the cylindrical pillar, diameter d = 50 cm = 0.5 m and height h = 3.5 m.
Curved surface area of the cylindrical pillar
= πdh
= \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 5.5 m²
Cost of painting 1 m² area = ₹ 12.50
∴ Cost of painting 5.5 m2 area = ₹ (12.50 x 5.5)
= ₹ 68.75
Thus, the cost of painting the curved surface of the pillar is ₹ 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.
Answer:
For the given cylinder, radius r = 0.7 m and
curved surface area = 4.4 m².
Curved surface area of a cylinder = 2πrh
∴ 4.4 m² = 2 × \(\frac{22}{7}\) × 0.7m × h
∴ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\)m
∴ h = 1 m
Thus, the height of the cylinder is 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.
Answer:
A circular well means a cylindrical well. For the cylindrical well, diameter d = 3.5 m and height (depth) h = 10 m.
(i) Curved surface area of the well
= πdh
= \(\frac{22}{7}\) × 3.5 × 10 m²
= 110 m²

(ii) Cost of plastering 1 m² region = ₹ 40
∴ Cost of plastering 110 m2 region
= ₹ (40 × 110)
= ₹ 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:
For the cylindrical pipe, diameter d = 5 cm = 0.05 m and height (length) h = 28 m.
The radiation surface in the system is the •curved surface of the pipe.
Hence, we find the curved surface area of the cylindrical pipe.
Curved surface area of the cylindrical pipe
= πdh
= \(\frac{22}{7}\) × 0.05 × 28 m²
= 4.4 m²
Thus, the total radiating surface in the system is 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.
Answer:
For the closed cylindrical tank, diameter d = 4.2 m, hence radius
r = \(\frac{4.2}{2}\) = 2.1 m and height h = 4.5 m.

(i) Curved surface area of the cylindrical tank
= 2 πrh
= 2 × \(\frac{22}{7}\) × 2.1 × 4.5 m²
= 59.4 m²

(ii) Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5) m²
= 13.2 × 6.6 m²
= 87.12 m²
Suppose, x m² steel was used for making the tank. But during production, \(\frac{1}{12}\) of the steel was wasted.
∴ Actual quantity of steel used = \(\frac{11}{12}\)x m².
Hence, \(\frac{11}{12}\)x = 87.12
∴ x = \(\frac{8712}{100} \times \frac{12}{11}\)
∴ x = 95.04 m²
Thus, the quantity of steel actually used during the preparation of the tank is 95.04 m².

Question 10.
In the given 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.

Answer:
The shape of the decorative cloth will be cylindrical.
For the cylinder of cloth, diameter d = 20 cm and height h = 30 cm + 2.5 cm + 2.5 cm = 35 cm.
Curved surface area of the cylinder of cloth
= πdh
= \(\frac{22}{7}\) × 20 × 35 cm²
Thus, 2200 cm² cloth is required for covering the lampshade.

Question 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 cardboards. If there were 35 competitors, how much cardboard was required to be bought for the competition ?
Answer:
The cylindrical penholders to be made have base but open at the top. Thus, to prepare a penholder, the area of the cardboard required will be given by the curved surface area of the cylinder and the area of base.

For cylindrical penholder, radius r = 3 cm and height h = 10.5 cm.
Area of cardboard required for 1 penholder
= Curved surface area of cylinder + Area of base
= 2πrh + πr²
= πr (2h + r)
= \(\frac{22}{7}\) × 3(2 × 10.5 + 3) cm2
= \(\frac{66 \times 24}{7}\) cm²
∴ Area of the cardboard required for 35 penholders
= 35 × \(\frac{66 \times 24}{7}\) cm²
= 7920 cm²
Thus, 7920 cm² cardboard was required to be bought for the competition.