Question

The diameter of a road roller 1m40cm long is 80cm .If it take 600 revolution to level a playground. Find the cost of leveling the ground at ₹3 per sq. meter
Hint : revolution ( ek complete rotation) or ek complete rotation kiske brabar hota uske perimeter ke​

Answer 1

Answer:

Rupees 6336/-

Step by step explanation:

Road Roller(Cylinder)

D=80cm

Therefore R=40cm

H= 1m40cm= 140cm

Therefore CSA of cube = 2πrh

=2×22/7×40×140

=2×22×40×20

=35200cm^2

=352m^2

Total number of revolutions taken by roller= 600

Therefore =352m^2×600

=2112m^2

Cost of levelling the ground per m^2 = rupees 3/-

= 2112m^2 × rupees 3/-

= Rupees 6336/-

<<HOPE THIS ANSWER HELPS YOU>>

Please Thank me and Vote me...