Q.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. (Assume π = 22 / 7)
Solution: Height of cylinder, h = 14cm
Let the diameter of the cylinder be d
Curved surface area of cylinder = 88 cm^{2}
We know that, formula to find Curved surface area of cylinder is 2πrh.
So 2πrh = 88 cm^{2} (r is the radius of the base of the cylinder)
2 × (22 / 7) × r × 14 = 88 cm^{2}
2r = 2 cm
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Q.2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22 / 7
Solution: Let h be the height and r be the radius of a cylindrical tank.
Height of cylindrical tank, h = 1m
Radius = half of diameter = (140 / 2) cm = 70cm = 0.7m
Area of sheet required = Total surface are of tank = 2πr(r + h) unit square
= [2 × (22 / 7) × 0.7(0.7 + 1)] = 7.48
Therefore, 7.48 square meters of the sheet are required.
Q.3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area (Assume π = 22 / 7)
Solution: Let r_{1} and r_{2} Inner and outer radii of cylindrical pipe
r_{1} = 4 / 2 cm = 2 cm
r_{2} = 4.4 / 2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i) curved surface area of outer surface of pipe = 2πr1h
= 2 × (22 / 7) × 2 × 77 cm^{2}
= 968 cm^{2}
(ii) curved surface area of outer surface of pipe = 2πr_{2}h
= 2 × (22 / 7) × 2.2 × 77 cm^{2}
= (22 × 22 × 2.2) cm^{2}
= 1064.8 cm^{2}
(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
= 2r_{1}h + 2r_{2}h + 2π(r_{1}^{2}  r_{2}^{2})
= 9668 + 1064.8 + 2 × (22 / 7) × (2.22  22)
= 2031.8 + 5.28
= 2038.08 cm^{2}
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm^{2}.
Q.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 m^{2}? (Assume π = 22 / 7)
Solution: A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84 / 2) cm = 42 cm
Now, CSA of roller = 2πrh
= 2 × (22 / 7) × 42 × 120
= 31680 cm^{2}
Area of field = 500 × CSA of roller
= (500 × 31680) cm^{2}
= 15840000 cm^{2}
= 1584 m^{2}.
Therefore, area of playground is 1584 m^{2}.
Q.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 m^{2}. (Assume π = 22 / 7)
Solution: Let h be the height of a cylindrical pillar and r be the radius.
Given:
Height cylindrical pillar = h = 3.5 m
Radius of the circular end of pillar = r = diameter / 2 = 50 / 2 = 25cm = 0.25m
CSA of pillar = 2πrh
= 2 × (22 / 7) × 0.25 × 3.5
= 5.5 m^{2}
Cost of painting 1 m^{2} area = Rs. 12.50
Cost of painting 5.5 m^{2} area = Rs (5.5 × 12.50)
= Rs.68.75
Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m^{2} is Rs 68.75.
Q.6. Curved surface area of a right circular cylinder is 4.4 m^{2}. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22 / 7)
Solution: Let h be the height of the circular cylinder and r be the radius.
Radius of the base of cylinder, r = 0.7m CSA of cylinder = 2πrh
CSA of cylinder = 4.4m^{2}
Equating both the equations, we have
2 × (22 / 7) × 0.7 × h = 4.4
Or h = 1
Therefore, the height of the cylinder is 1 m.
Q.7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m^{2}.
(Assume π = 22 / 7)
Solution: Inner radius of circular well, r = 3.5 / 2m = 1.75m
Depth of circular well, say h = 10m
(i) Inner curved surface area = 2πrh
= (2 × (22 / 7) × 1.75 × 10) = 110
Therefore, the inner curved surface area of the circular well is 110 m^{2}.
(ii) Cost of plastering 1 m^{2} area = Rs.40
Cost of plastering 110 m^{2} area = Rs (110 × 40)
= Rs.4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.
Q.8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22 / 7)
Solution: Height of cylindrical pipe = Length of cylindrical pipe = 28m
Radius of circular end of pipe = diameter / 2 = 5 / 2 cm = 2.5cm = 0.025m
Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder
= 2 × (22 / 7) × 0.025 × 28 m^{2}
= 4.4m^{2}
The area of the radiating surface of the system is 4.4m^{2}.
Q.9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.
(ii) How much steel was actually used, if 1 / 12 of the steel actually used was wasted in making the tank. (Assume π = 22 / 7)
Solution: Height of cylindrical tank, h = 4.5m
Radius of the circular end , r = (4.2 / 2)m = 2.1m
(i) the lateral or curved surface area of cylindrical tank is 2πrh
= 2 × (22 / 7) × 2.1 × 4.5 m^{2}
= (44 × 0.3 × 4.5) m^{2}
= 59.4 m^{2}
Therefore, CSA of tank is 59.4 m^{2}.
(ii) Total surface area of tank = 2πr(r + h)
= 2 × (22 / 7) × 2.1 × (2.1 + 4.5)
= 44 × 0.3 × 6.6
= 87.12 m^{2}
Now, Let S m^{2} steel sheet be actually used in making the tank.
S(1  1 / 12) = 87.12 m^{2}
This implies, S = 95.04 m^{2}
Therefore, 95.04m^{2} steel was used in actual while making such a tank.
Q.10. In fig. 13.12, 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. (Assume π = 22 / 7)
Solution: Say h = height of the frame of lampshade, looks like cylindrical shape
r = radius
Total height is h = (2.5 + 30 + 2.5) cm = 35cm and
r = (20 / 2) cm = 10cm
Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh
= (2 × (22 / 7) × 10 × 35) cm^{2}
= 2200 cm^{2}
Hence, 2200 cm^{2} cloth is required for covering the lampshade.
Q.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 cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π = 22 / 7)
Solution: Radius of the circular end of cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh + πr^{2}
= 2 × (22 / 7) × 3 × 10.5 + (22 / 7) × 32 = 1584 / 7
Therefore, Area of cardboard sheet used by one competitor is 1584 / 7 cm^{2}
So, Area of cardboard sheet used by 35 competitors = 35 × 1584 / 7 = 7920 cm^{2}
Therefore, 7920 cm^{2} cardboard sheet will be needed for the competition.
1402 docs679 tests

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