Question

Diameter and length of roller are 84 cm and 120 cm respectively. In how many revolutions, can the roller level the playground of area 1,584m2 ?

Answer 1

Area covered by roller in 1 revolution = C.S.A
Radius is 42 cm.
Now C.S.A. of roller = 2×pi×r×h or length
= 2×22/7×42×120
= 2×22×6×120
= 31680 cm^2
No. of revolutions = area to be covered / C.S.A.
= 1584 × 100 × 100 / 31680
= 500

Hence , the roller can complete 500 rounds to level the playground

Answer 2

${\bold{\underline{\boxed{Answer = 500 \:revolutions}}}}$

Given : we know that the shape of the roller is cylindrical.

Diameter of the cylindrical roller = 84 cm

Radius of the roller =$\dfrac{diameter}{2}=\dfrac{84}{2}=42\:cm$

Height of the cylindrical roller = 120 cm

Area of the playground is 1,584 m²

CSA of the roller = 2πrh

= $2\times\dfrac{22}{7}\times42\times120$

= $2\times\dfrac{22}{\cancel{7}}\times\cancel{42}\times120$

= 44 × 6 × 120 = 31,680 cm²

Therefore, Area covered = 1584 sq m

$(\because1sq\:m=10,000sq\:cm)$

So 1,584 × 10,000 = 15,840,000 cm²

Total number of revolutions to cover the area of 15,840,000 sq cm

= $\sf\cancel\dfrac{15,840,000}{31,680}=500\:revolutions.$