Question

A cylindrical roller with diameter 77 cm and height 1 m is used to level a rectangular park with dimensions22 × 11 . Find the number of revolutions required to level the park.

Answer 1

Answer:  The number of revolutions required to level the park is 100.

Step-by-step explanation:

Given : Diameter of roller = 77 cm

Then radius of roller : r = $\dfrac{77}{2}\ cm$

Height = 1 m = 100 cm

Curved surface area of roller = $2\pi rh= 2\times\dfrac{22}{7}\times\dfrac{77}{2}\times100=24200\ cm^2$

Dimensions of park = 22 × 11 m

Then Area of park = $22\times11\ m^2= 232\ m^2=232\times100\times100 \ cm^2$

$=2420000\ cm^2$

Now , the number of revolutions required to level the park = (Area of park) ÷ (Curved surface area of roller )

$2420000\div 24200=100$

Hence, the number of revolutions required to level the park is 100.