Answer :

Given: length of cylindrical roller, i.e., height of the cylinder, h = 2.5 m

Radius of cylindrical roller, r = 1.75 m

The cylindrical roller will roll on the road only from its curved surface.

So, curved surface area of cylindrical roller, CSA = 2πrh

⇒ CSA = 2 × 22/7 × 1.75 × 2.5 = 27.5 m^{2}

⇒ area of road covered in 1 revolution = 27.5 m^{2}

And given that, total area of road covered = 5500 m^{2}

So, number of revolutions made by road roller to cover 5500 m^{2} = 5500/27.5 = 200

**Thus, it made 200 revolutions.**

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