Question

1. Diameter and length of a roller are 84 cm and 120 cm respectively. In how many revolutions, can the roller level the playground of area 1,584 m??​

Answer 1

Answer:

$\bigstar{\bold{Number\:of\:revolutions=500}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Diameter of the roller = 84 cm = 0.84 m
• Length of the roller = 120 cm = 1.2 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Number of the revolutions required to level an area of 1584 m²

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we have to find the curved surface area of the roller.

→ Here the roller is in the shape of a cylinder.

→ CSA of a cylinder is given by,

  CSA of a cylinder = 2 π r h

→ where r = d/2 = 0.42 m

  and h = length = 1.2

→ Subtituting the data,

  CSA of the roller = 2 × 22/7 × 0.42 × 1.2

  CSA of the roller = 2 × 22 × 0.06 × 1.2

  CSA of the roller = 44 × 0.072

  CSA of the roller = 3.168 m²

→ Now given that the area of the playground is 1584 m²

→ Hence,

  Number of revolutions = Area of playground/CSA of the roller

→ Substitute the data,

  Number of revolutions = 1584/3.168

  Number of revolutions = 500

→ Hence the number of revolutions required is 500

$\boxed{\bold{Number\:of\:revolutions=500}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The CSA of a cylinder is given by

   CSA of a cylinder = 2 π r h

→ The TSA of a cylinder is given by

  TSA of a cylinder = 2 π r ( r + h)

→ The volume of a cylinder is given by

   Volume of the cylinder = π r² h

Answer 2

Answer:

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Step-by-step explanation:

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