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Question
The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square metre.
Solution
Given that,
Diameter of the roller=84cm=0.84m.
Length of the roller=1.5m
Radius of the roller =`D/2=0.84/2=0.42.`
Area covered by the roller on one revolution = covered surface area of roller
Curved surface of roller= `2pirh=2xx22/7xx0.42xx1.5`
=`0.12xx22xx1.5m^2`
Area of the playground=`"100xxArea covered by roller in one revolution"`
=`(100xx0.12xx22xx1.5)m^2`
=`396m^2`
Now ,
Cost of leveling `1 m^2=50P=50/100⇒Re=1/2rs`
Cost of leveling `396m^2=1/2xx396=Rs.198`
Hence, cost of leveling `396m^2 is 198`
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