The diameter of a road roller, 120 cm long is 84 cm. If it takes 500 complete revolutions to level a play ground, find the cost of levelling it at the rate of Rs. 2 per square metre.

Outer diameter (D) = 25 cm
R = 25/2 = 12.5 cm
Inner diameter(d) = 23 cm
r = 23/2 = 11.5 cm
Height = 20 cm
Total surface area = 2*22/7*r (R+r)(h+R-r)
​= 2 * 22/7 * (24) (21)
= 44 * 24 * 3
= 3168 cm^2
Therefore t.s.a is 3168​cm^2
and area covered bythe roller in 500 revolutions = 3168 * 500
= 1584000 sq. cm
= 1584 sq. m
1 m^2 cost = Rs2
So​1584 sq. m cost = [(2*1584) Rs.]
= Rs. 3618
hope this helps.