length and radius of a roller is 14 m and 7m respectively.i) What is the curved surface area of the roller? ii) If the length and breadth of the ground is 400 m and 120 m respectively, how many complete revolutions will it take to level the ground?
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Step-by-step explanation:
Given that r=42cm and h=120cm
Area covered by the roller in one revolution=Curved Surface Area of the road roller.
=2πrh
=2×
7
22
×42×120
31680cm
2
Area covered by the roller in 500 revolutions =31680×500
=15840000cm
2
=
10000
15840000
=1584m
2
[10000cm
2
=1sq.m]
Cost of levelling per 1sq.m = Rs.
100
75
Thus, cost of levelling the play ground =
100
1584×75
= Rs.1188.
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