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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?
Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
= 2πrh + πr2
Area of cardboard sheet used by 1 competitor 
Area of cardboard sheet used by 35 competitors
=
= 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be bought.
The curved surface area of a right circular cylinder of height 14 cm is . Find the diameter of the base of the cylinder.
Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
⇒ 2πrh = 88 cm2 (r is the radius of the base of the cylinder)
⇒ πdh = 88 cm2 (d = 2r)
⇒
⇒ d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
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?
Height (h) of cylindrical tank = 1 m
Base radius (r) of cylindrical tank
Therefore, it will require 7.48 m2 area of sheet.
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.
Inner radius 
of cylindrical pipe 
Outer radius 
of cylindrical pipe 
Height (h) of cylindrical pipe = Length of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe
(ii) CSA of outer surface of pipe 
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
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
It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller = 
CSA of roller = 2πrh
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 1584 m2
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
Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar = 
= 0.25 m
CSA of pillar = 2πrh
Cost of painting 1 m2 area = Rs 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs 68.75
Therefore, the cost of painting the CSA of the pillar is Rs 68.75.
Curved surface area of a right circular cylinder is 4.4 . If the radius of the base of the cylinder is 0.7 m, find its height.
Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
2πrh = 4.4 m2
h = 1 m
Therefore, the height of the cylinder is 1 m.
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 .
Inner radius (r) of circular well
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
= (44 × 0.25 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
Cost of plastering 1 m2area = Rs 40
Cost of plastering 110 m2area = Rs (110 × 40)
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
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.
Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = 
= 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2πrh
= 4.4 m2
The area of the radiating surface of the system is 4.4 m2.
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 of the steel actually used was wasted in 12 making the tank.
Height (h) of cylindrical tank = 4.5 m
Radius (r) of the circular end of cylindrical tank = 
(i) Lateral or curved surface area of tank = 2πrh
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2πr (r + h)
= (44 × 0.3 × 6.6) m2
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
Therefore, 95.04 m2 steel was used in actual while making such a tank.
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.
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius (r) of the circular end of the frame of lampshade = 
Cloth required for covering the lampshade = 2πrh
= 2200 cm2
Hence, for covering the lampshade, 2200 cm2 cloth will be required.
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