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 : Given,
Curved surface area, C. S. A = 88 cm²
and height, h = 14 cm
We know,
Curved surface area of the cylinder = 2πrh
where, r = radius of the base
and h = height of the cylinder
or, C. S. A = 2πrh ……(1)
Substitute, C.S.A = 88 cm² h = 14 cm and π = 22/7 in (1)
88 = 2 × 22/7 × r × 14
or, r = ( 88 × 7)/(2 × 22 × 14) = 1 m
Then, diameter of the base = 2 r = 2 × 1 = 2 cm.
Hence, the diameter of the base is 2 cm.
2). It is required to make a closed cylindrical tank of height 1 m and base of diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
Solution : Given,
Height of the cylindrical tank, h = 1 m
and base diameter = 140 cm
∴ Radius, r = diameter/2 = 140/2 = 70 cm
or, r = 70/100 m = 0.7 m
Now,
Area of metal sheet = total surface area of the cylinder
∵ total surface area of the cylinder = 2πr ( r + h)
= 2 × 22/7 × 0.7 × ( 1 + 0.7 )
= 2 × 22/7 × 0.7 × 1.7 m² = 7.48 m²
Hence, the area of metal sheet is 7.48 m²
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. Find its :
i) inner curved surface area. ii) outer curved surface area. iii) total surface area.
SOLUTION : Let R and r be the outer and inner radii of the metal pipe and h be the length of the metal pipe.
Given that,
Outer diameter = 4.4 cm
Inner diameter = 4 cm,
and length of the metal pipe = 77 cm.
Then, outer radius, R = 4.4/2 = 2.2 cm [ ∵ radius = diameter/2]
inner radius, r = 4/2 = 2 cm, and h = 77 cm
i) Let S₁ = inner curved surface area of the metal pipe = inner curved surface area of cylinder.
∴ S₁ = 2πr h = 2 × 22/7 × 2 × 77 cm² = 968 cm².
ii) Let S₂ be the outer curved surface area of the metal pipe, then
S₂ = 2π R h = 2 × 22/7 × 2.2 × 77 cm²
= 44 × 2.2 × 11 cm² = 1064.8 cm²
iii) ∵ Total surface area = Inner curved surface area + outer curved surface area + area of two bases
or, total surface area = S₁ + S₂ + 2 ( πR² - π r² )
= 968 + 1064.8 + 2 × 22/7 [ (2.2)² - 2² ]
= 968 + 1064.8 + 44/7 × 0.84
= 968 + 1064.8 + 5.28 cm² = 2038.08 cm²
Hence, the i) inner surface area is 968 cm² , ii) outer surface area is 1064.8 cm² and iii) total surface area is 2038.08 cm²
4). The diameter of a roller 120 cm long is 84 cm. If it take 500 complete revolutions to level a playground, determine the area of playground in m² ?
SOLUTION : Given,
Length of the roller = 120 cm,
and diameter of the roller = 84 cm
∵ Roller is a right circular cylinder.
∴ radius of the cylinder, r = diameter/2 = 84/2
or, r = 42 cm
Number of revolutions = 500
and height of the cylinder, h = 120 cm
Now,
Curved surface area of the cylinder = 2πrh
= 2 × 22/7 × 42 × 120 cm² = 31680 cm²
= 3.1680 m²
∴ Area covered by the roller is 500 revolutions
= 500 × 3.1680 m² = 1584 m²
Hence, the area of the playground in 500 revolutions is 1584 m².
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 : Given,
Diameter of the cylindrical pillar = 50 cm
Height of the pillar = 3.5 m
and cost of painting is 12.50 per m²
Now,
Radius of the cylinder, r = diameter/2
r = 50/2 = 25 cm
= 0.25 m
Height of the cylinder, h = 3.5 m
∴ Curved surface area of the cylinder = 2πrh
= 2 × 22/7 × 0.25 × 3.5 m² = 5.5 m²
∵ Cost of painting in 1 m² = Rs 12.50
∴ Cost of painting in 5.5 m² = Rs 12.50 × 5.5
= Rs 68.75
Hence, the cost of painting is Rs 68.75
