A road roller takes 750 complete revolutions to move once over to level a road.

Question:

A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84 cm and its length is 1 m.

Solution:

Diameter $=84 \mathrm{~cm}$

i.e., radius $=42 \mathrm{~cm}$

Length $=1 \mathrm{~m}=100 \mathrm{~cm}$

Now, lateral surface area $=2 \pi \mathrm{rh}=2 \times \frac{22}{7} \times 42 \times 100=26400 \mathrm{~cm}^{2}$

$\therefore$ Area of the road = lateral surface area $\times$ no. of rotations $=26400 \times 750=19800000 \mathrm{~cm}^{2}=1980 \mathrm{~m}^{2}$

Leave a comment