The diameter of roller 1.5 long is 84 cm. If it takes 100 revolution to level a play ground ,find the cost of 50 paise per sqare meter
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Given that,
Diameter of the roller = 84 cm = 0.84 m
Length of the roller = 1.5 m
Radius of the roller = D/2 = 0.84/2 = 0.42
Area covered by the roller on one revolution = covered surface area of roller
Curved surface of roller = 2πrh = 2 x 22/7 x 0.42 x 1.5
= 0.12 x 22 x 1.5 m2
Area of the playground = 100 x Area covered by roller in one revolution
= (100 x 0.12 x 22 x 1.5) m2
= 396 m2
Now,
Cost of levelling
= 396 × 50/100
= RS.198
therefore the cost of levelling the playground is rupees 198.
Diameter of the roller = 84 cm = 0.84 m
Length of the roller = 1.5 m
Radius of the roller = D/2 = 0.84/2 = 0.42
Area covered by the roller on one revolution = covered surface area of roller
Curved surface of roller = 2πrh = 2 x 22/7 x 0.42 x 1.5
= 0.12 x 22 x 1.5 m2
Area of the playground = 100 x Area covered by roller in one revolution
= (100 x 0.12 x 22 x 1.5) m2
= 396 m2
Now,
Cost of levelling
= 396 × 50/100
= RS.198
therefore the cost of levelling the playground is rupees 198.
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