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Question
A road roller has a diameter of 84 cm and it's width is 1.2 m. If it takes 500 complete revolution to level a ground. Find area of the ground. Also find the cost of leveling the ground at the rate of 50 paise per metre square
A road roller has a diameter of 84 cm and it's width is 1.2 m. If it takes 500 complete revolution to level a ground. Find area of the ground. Also find the cost of leveling the ground at the rate of 50 paise per metre square
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Solution
Diameter of roller = 84 cm
Radius = 84/2 = 42 cm or 0.42 m
Width or height = 1.2 m
Roller is in the shape of a cylinder.
Area of roller = Curved surface area of cylinder
Curved surface area of cylinder = 2πrh
⇒ 2*22/7*0.42*1.2
⇒ 22.176/7
= 3.168 m²
Area of 1 revolution = Area of roller = 3.168 m²
Area of 500 revolution = Area of playground = 3.168 × 500
= 1584 m²
So, the area of the playground is 1584 m²
Cost of levelling the playground should be Rs. 50 per sq m instead of wrongly written 50 % per square meter.
Total cost of levelling the playground = 50 *1584
= Rs, 79200
Radius = 84/2 = 42 cm or 0.42 m
Width or height = 1.2 m
Roller is in the shape of a cylinder.
Area of roller = Curved surface area of cylinder
Curved surface area of cylinder = 2πrh
⇒ 2*22/7*0.42*1.2
⇒ 22.176/7
= 3.168 m²
Area of 1 revolution = Area of roller = 3.168 m²
Area of 500 revolution = Area of playground = 3.168 × 500
= 1584 m²
So, the area of the playground is 1584 m²
Cost of levelling the playground should be Rs. 50 per sq m instead of wrongly written 50 % per square meter.
Total cost of levelling the playground = 50 *1584
= Rs, 79200
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