The diameter of a roller is 84, cm and its le | vedclass

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(C) The roller is cylindrical in shape.Height $(h)$ of the cylindrical roller = Length of the roller = $120\, cm$.Radius $(r)$ of the circular end of

Vedclass · Accepted Answer

$1584$

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(C) The roller is cylindrical in shape.Height $(h)$ of the cylindrical roller = Length of the roller = $120\, cm$.Radius $(r)$ of the circular end of the roller = $\frac{84}{2}\, cm = 42\, cm$.Curved Surface Area $(CSA)$ of the roller = $2\pi rh$.$CSA = 2 	imes \frac{22}{7} 	imes 42 	imes 120\, cm^2$.$CSA = 2 	imes 22 	imes 6 	imes 120\, cm^2 = 31680\, cm^2$.The area of the playground is equal to the total area covered in $500$ revolutions.Area of the playground = $500 	imes CSA$ of the roller.Area = $500 	imes 31680\, cm^2 = 15840000\, cm^2$.Since $1\, m^2 = 10000\, cm^2$,we convert the area to $m^2$:Area = $\frac{15840000}{10000}\, m^2 = 1584\, m^2$.

The diameter of a roller is $84\, cm$ and its length is $120\, cm$. It takes $500$ complete revolutions to move once over to level a playground. Find the area of the playground in $m^2$. $\left[\text{Assume } \pi = \frac{22}{7}\right]$ (in $, m^2$)

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