Question

A roller of height 10 m and radius 0.35 m is used to level a lawn. Find the area (in sq. m) it can level in one full rotation.

Answer 1

hope my answer helps you

Answer 2

⇒ Final Answer:

22 m²

⇒ Given:

Height of the roller = 10 m

Radius of the roller = 0.35 m

⇒ To Find:

The area the roller can level in one full rotation.

⇒ Analysis:

A roller that is used to level a ground in cylindrical in shape. Also while levelling, only the curved part of the roller comes in contact with the ground. Hence, in order to find the area that gets covered in one full rotation, we have to find the CSA of the cylindrical roller.

⇒ Formula to be used:

→ $\boxed{\sf{CSA\:of\:Cylinder=2\pi\:r\:h}}$

⇒ Solution:

Height = 10 m

Radius = 0.35 m

CSA of cylinder = 2πrh

$\sf{=2\times\pi\times0.35\times10}$

$\sf{2\times\dfrac{22}{7}\times0.35\times10}$

$\sf{22\:m^2}$

Hence the required answer is 22 m².

The cylindrical roller's CSA is 22 m².

Knowledge Bytes:

→ Cylinder

A cylinder is a 3D figure which has 3 curved faces. It has 2 edges and 0 vertices.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{r}}\put(9,17.5){\sf{h}}\end{picture}$

→ Formulae related to cylinder:

✳ $\sf{CSA=2\pi\:r\:h}$

✳ $\sf{TSA=2\pi\:r(r+h)}$

✳ $\sf{Volume=\pi\:r^2h}$