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ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.1 in Chapter 17 - Mensuration
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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.
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 m2
Area of the play ground = l×b
=120×44
= 5280 m2
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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