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Answer :
Given,diameter of the road roller=84cm and length=1m.Also revolution=750
We have radius of roller r=\(\frac{84}2=42cm=0.42m\)
So, curved surface area of the roller=\(2\pi rh\)
\(=2\times \frac{22}7 \times 0.42 \times 1\)
\(=22.64 m^2\)
∴area levelled by the roller in 750 revolutions:
\(\qquad=(750\times 2.64)=1980m^2\)
Hence, the area of the road levelled=\(1980m^2\)