Question

the diametre of a roller 120 cm long is 84 cm . If its take 500 revolution to level the play ground determine the cost of leveling it at the rate of 45 paise per sq metre


plz tell me fast ​

Answer 1

Answer:

Area of 1 revolution of roller

$2\pi \: rh = 2 \times \frac{22}{7} \times 60 \times 84 \\ = 31,680 \: {cm}^{2} = 3.168 \: {m}^{2}$

Area of 500 revolution of roller

31680*500 = 15,840,000 cm^2

$= 1584 \: {m}^{2}$

Total cost of levelling,

= 1584*0.45 = Rs 712.8