Diameter of a roller = 140 cm
Radius of the roller :
[tex] =\tt \frac{diameter}{2} [/tex]
[tex] =\tt \frac{140}{2} [/tex]
[tex] = \tt70 \: cm[/tex]
Thus, the radius of the roller = 70 cm
Height of roller = 2 m = 2×100 cm = 200 cm
Curved surface area of roller = 1 revolution
The curved surface area of this roller :
[tex] =\tt 2\pi rh[/tex]
[tex] = \tt2 \times 3.14 \times 70 \times 200[/tex]
[tex] =\tt87920 \: {cm}^{2} [/tex]
Thus, the curved surface area of the roller = 87920 cm²
Number of revolutions that needs to be done to level the ground = 350
Distance the roller will travel to get leveled :
[tex] =\tt 87920 \times 350[/tex]
[tex] = \tt43960000 \: cm[/tex]
[tex] = \tt \frac{43960000}{100} \: m[/tex]
[tex] = \tt439600 \: m[/tex]
Thus, to level the playground the distance the roller should cover = 539600 m
Cost of covering one square meter of land = 75 Paise
Cost of covering 539600 m of land :
[tex] =\tt 539600 \times \frac{75}{100} \: Rs[/tex]
[tex] = \tt \frac{539600 \times 75}{100} \: Rs[/tex]
[tex] =\tt \frac{40470000}{100} \: Rs[/tex]
[tex]\color{plum} = \tt\bold{404700 \: rupees}[/tex]
▪︎The cost of leveling the playground = 404700 Rs
