Answer:
Diameter of road roller = 84 cm r = \[\frac{84}{2}=42\]cm = 0.42 m Length of road roller = 1 m The shape of road roller is cylindrical, then 1 revolution to move once over level the road = 1 surface area of road roller \[=\text{ }2\pi rh\] \[=2\times \frac{22}{7}\times 0.42\times 1\] \[=\text{ }2.64\text{ }{{m}^{2}}\] The area of 750 revolutions to level the road by a road roller \[=750\times 2.64\] \[=\text{ }1980\text{ }{{m}^{2}}\]
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