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Question 92
A field roller, in the shape of a cylinder, has a diameter of 1 m and length of $$1\frac{1}{4}$$ m. If the speed at which the roller rolls is 14 revolutions per minute, then the maximum area (in m$$^2$$) that it can roll in 1 hour is: (Take $$\pi = \frac{22}{7}$$)
Solution
Radius = 1/2 m
Length = $$1\frac{1}{4} = \frac{5}{4}$$
1 revolution = $$2 \pi r l = 2 \times \frac{22}{7} \times \frac{1}{2} \times \frac{5}{4} = \frac{55}{14}$$
Maximum rolling area in 1 hr = $$\frac{55}{14} \times 14 \times 60$$ = 3300
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