The diameter of a road roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling it at the rate of 30 paise per sq metre
Answers
Answer:
The roller is cylindrical in shape and hence it is considered as a right circular cylinder. In one revolution, the area covered will be the curved surface area of the roller.
Since it takes 500 complete revolutions to move once over to level a playground, the area of the playground will be equal to 500 times the curved surface area of the roller.
Let the radius and height of the cylinder are 'r' and 'h' respectively.
Curved Surface Area of the cylinder = 2πrh
Length of the roller, h = 120 cm
Radius of the roller, r = 84/2 cm = 42 cm
Curved Surface Area of the roller = 2πrh
= 2 × 22/7 × 42 cm × 120 cm
= 31680 cm²
Area of the playground = Area leveled by the cylinder in 500 revolutions
= 500 × 31680 cm²
= 15840000 cm²
= 15840000/10000 m² [Since 1cm² = 1/10000 m²]
= 1584 m²
Thus, area of the playground = 1584 m².
Answer:
diameter of the cylinder = 120cm
radius = 60cm
height of the cylinder = 84cm
curved surface of the area
= 2πrh
= 2 * (22/7) * 60 * 84
= 2 * 22 * 60 * 12
= 31,680cm²
area covered for 1 revolution = 31,680cm²
area covered for 500 revolutions
= 31680 * 500
= 1,58,40,000cm²
= 1,584m²
cost for levelling = 1,584 * 30 = 47,520 paise = ₹475.2
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