PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2
PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2
PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2
Note: Assume π = 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
∴ 88×7/2×22×14 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 = diameter/2
= 140/2 cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × 227 × 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 = diameter/2 = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × 22/7 × 2 × 77 cm2
= 968 cm2
Thus, the inner curved surface area is 968 cm2.
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
= 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 × 22/7 × 0.7m × h
∴ h = 4.4×7/2×22×0.7m
∴ 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
= 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
= 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 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 = 4.2/2 = 2.1 m and height h = 4.5 m.
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
= 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.
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