Question

A roller of diameter 0.9 m. and length 1.8 m. is used to press the ground. Find the area of ground pressed by it in 500 revolutions. (Given $\mathrm{\pi }=3.14$)

Answer 1

Radius of the cylindrical roller = $\frac{\mathrm{Diameter}}{2}=\frac{0.9}{2}=0.45\mathrm{m}$
Length of the cylindrical roller = 1.8 m

Surface area covered by the roller in 1 revolution = 2$\mathrm{\pi }$rh
= 2 × 3.14 × 0.45 × 1.8
= 5.087 m²
Thus, we have:
Surface area covered by the roller in 500 revolutions = 5.087 × 500
= 2543.5 m²