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The diameter of a roller 120 cm long is 84cm . If it takes 500 complete revolutions to level a playground , determine the cost of levelling it at the rate of 30 paise per square metre .

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Solution : Clearly , the roller is a right circular cylinder of height h = 120 cm radius of its base =r= 42 cm . 

 	herefore            Area covered by the roller inone revolutions = Curved surface area of the roller 

                                                                                       = 2	imes frac227	imes 42	imes 120= 31680cm^2

So , Area covered by the roller is 500 revolutions  =(31680	imes500)cm^2

                                                                                 =frac(31680	imes500)100	imes 100m^2=1584m^2

 Hence , cost of levelling the playground          = Rs 1584 	imes frac30100= Rs 475.20

Posted by

Deependra Verma

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