Question

diameter of roadroller of length 120 is 84cm.If it takes 100 revolution ,then how much cost for levelling at 10₹per square meter​

Answer 1

Answer:

Given that r=42cm and h=120cm

Area covered by the roller in one revolution=Curved Surface Area of the road roller.

=2πrh

=2×722×42×120

31680cm2

Area covered by the roller in 500 revolutions =31680×500

=15840000cm2

=1000015840000=1584m2[10000cm2=1sq.m]

Cost of levelling per 1sq.m = Rs.10075

Thus, cost of levelling the play ground =1001584×75= Rs.1188.