Question

The diameter of a road roller of length120cm is 84cm. If it takes 500 complete revolutions to level a playground, then find the cost of leveling it at the cost of 75 paise persq.m

Answer 1

Answer:

Step-by-step explanation:

r=42cm (84/2) :  h=120cm

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

=2πrh

=2×  22/7 ×42×120

=31680cm  

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

=15840000cm   =1584m  

Cost of levelling per 1sq.m = 75 p = Rs. 75/100

Thus, cost of levelling the play ground =  1584×75 /100

                                                                  = Rs.1188

Hope this answer helps u.

PLS MARK AS THE BRAINLIEST!!!!!!!!!

​