Question

7. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete
revolutions to roll once over the play ground to level. Find the area of the play ground
in m^2​

Answer 1

${\huge{\underline{\underline{\sf{\maltese\;Question:-}}}}}$

Q) The diameter of a roller is 84 cm and it's length is 120 cm . It takes 500 complete revolutions to roll once over the play ground to level . Find the area of the play ground in m² .

${\huge{\underline{\underline{\sf{\maltese\;Answer:-}}}}}$

☆ Given :

• Diameter of roller = 84 cm
• Length of the roller = 120 cm
• It takes 500 revolutions .

☆ To Find :

• Area of play ground in m²

☆ Solution :

• The roller is cylindrical .

Since we need the answer in m² , so , change cm into m .

• diameter = 84 cm = 0.84 m
• Length = 120 cm = 1.2 m

Radius = diameter/2 = 0.42 m

As we know that , when a roller will roll , it's CSA will be in the contact of the ground .

So ,

➝ Area covered in one revolution = CSA

➝ Area covered in 500 revolution = 500 × CSA = Area of Playground .

$\mapsto \rm \: area \: of \: playground = 500 \times csa$

$\mapsto \rm \: area \: of \: playground = 500 \times 2 \pi rh$

$\mapsto \rm \: area \: of \: playground = 1000 \times \frac{22}{ \cancel7 } \times \cancel{0.42 }\times 1.2 \: {m}^{2}$

$\mapsto \rm \: area = 1000 \times 22 \times 0.06 \times 1.2 \: {m}^{2}$

$\longrightarrow\purple{ \boxed{ \rm \: area \: of \: playground = 1584 \: {m}^{2} }}$

So ,

The area of Playground in m² would be : 1584 .

_________________________

☆ Know More :

• Volume of Cylinder = πr²h

• TSA of Cylinder = 2πr(h+r)

• CSA of Cylinder = 2πrh

• The unit of volume is in the form - unit³

• The unit of surface area is in the form - unit²

Answer 2

Answer:

The area of Playground in m² would be : 1584 .

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