Question

A roller has diameter of 0.7m and length of 1m. It is used to roll over the cricket pitch. The surface area of the cricket pitch is equal to 60m^2. How much revolution will roller make to move once over the pitch? (Take π = \(\frac{22}{7}\))

(a) 25

(b) 28

(c) 30

(d) 26

The question was posed to me in an interview.

The doubt is from Surface Area of a Right Circular Cylinder in division Surface Areas and Volumes of Mathematics – Class 9

Answer 1

The correct option is (b) 28

The explanation: The effective area that will be in contact with the pitch is the curved surface area of the cylinder.

Diameter of the roller = 0.7m

Hence, radius of the roller = 0.35m and height (length here) of the roller = 1m

We know that curved surface area of a cylinder = 2πr

= 2 * \(\frac{22}{7}\) * 0.35 * 1

= 2.2 m^2

Area of the cricket pitch = 60m^2

Let’s assume that the roller makes “n” revolutions to move once over the pitch.

Therefore, n * curved surface area of the roller = surface area of the pitch

n = \(\frac{Surface \,area \,of \,the \,pitch}{Curved \,surface \,area \,of \,the \,roller}\)

= \(\frac{60}{2.2}\)

= 27.27

≈ 28 revolutions.