A road roller of diameter 98 cm has a length of 125 cm. It makes 4000 complete revolutions in order to level a field. Find the cost (in Rs.) of leveling it at 32 Rs. per square meter.
Answers
Given : A road roller of diameter 98 cm has a length of 125 cm.
It makes 4000 complete revolutions in order to level a field.
To Find : the cost (in Rs.) of leveling it at 32 Rs. per square meter.
Solution:
One revolution = lateral surface area of roller
diameter 98 cm
=> radius r = 49 cm = 49/100 m
length of 125 cm
=> h = 125/100 m = 5/4 m
lateral surface area of roller = 2πrh
= 2 (22/7) *(49/100) (5/4)
= 77 * 5 /100
= 385/100 m²
4000 complete revolutions area covered = 4000 * 385/100
= 15400 m²
Cost = 32 * 15400 = 4,92,800 Rs
the cost (in Rs.) of leveling = Rs 492800
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Answer:
One revolution = lateral surface area of roller
diameter 98 cm
=> radius r = 49 cm = 49/100 m
length of 125 cm => h=125/100 m = 5/4 m
lateral surface area of roller = 2+rh = 2 (22/7) *(49/100) (5/4)
= 77 * 5/100
= 385/100 m²
4000 complete revolutions area covered = 4000* 385/100
= 15400 m²
Cost = 32* 15400 = 4,92,800 Rs
the cost (in Rs.) of leveling = Rs 492800.