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A road roller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

Answer:

**Solution:**

The area of any given cylinder's curved surface with a base radius of "r" and a height of "h" is referred to as the "curved surface area of a cylinder" (CSA)

Given diameter of the road roller, d = 0.7 m

Radius, r = d/2 = 0.7/2 = 0.35 m

Width, h = 1.2 m

Curved surface area of the road roller = 2π rh

=2 ×(22/7)×0.35×1.2

= 2.64 m^{2}

Area of the play ground = l×b

=120×44

= 5280 m^{2}

Number of revolutions = Area of play ground / Curved surface area

= 5280/2.64

= 2000

**Hence the road roller must take 2000 revolutions to level the ground.**

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