A cylindrical roller 2.5, m in length and 1.7 | vedclass

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(A) The length (height) of the cylindrical roller is $h = 2.5\, m$ and the radius is $r = 1.75\, m$.The area covered by the roller in one revolution i

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The length (height) of the cylindrical roller is $h = 2.5\, m$ and the radius is $r = 1.75\, m$.The area covered by the roller in one revolution is equal to its lateral surface area (curved surface area).Lateral Surface Area $= 2\pi rh = 2 	imes \frac{22}{7} 	imes 1.75 	imes 2.5$.$= 2 	imes \frac{22}{7} 	imes \frac{175}{100} 	imes 2.5 = 2 	imes 22 	imes 0.25 	imes 2.5 = 44 	imes 0.625 = 27.5\, m^{2}$.The total area covered on the road is $5500\, m^{2}$.Number of revolutions $= \frac{	ext{Total area covered}}{	ext{Area covered in one revolution}}$.Number of revolutions $= \frac{5500}{27.5} = 200$ revolutions.

$A$ cylindrical roller $2.5\, m$ in length and $1.75\, m$ in radius,when rolled on a road,was found to cover an area of $5500\, m^{2}$. How many revolutions did it make?

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