Question

Q.4) The length of a road roller is 3.5m and its radius is 1.4m. For leveling a ground 600 rotations
of the road roller were required. How much area of ground was leveled by the road roller? (4)
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Answer 1

Given

✭ Length of road roller is 3.5 m

✭ Radius of the roller is 1.4 m

✭ To level a ground it takes 600 rotation

To Find

◈ Area covered by it?

Solution

Concept

We shall first find the CSA of the cylinder because the top and the bottom of the the cylinder will not contribute in the levelling of the road then simply multiply the CSA with 600 to get our final answer!!

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We know that CSA of a cylinder is given by,

CSA = 2πrh

Substituting the given values,

➝ 2πrh

➝ 2 × 22/7 × 1.4 × 3.5

➝ 2 × 22 × 0.2 × 3.5

➝ CSA = 30.8 m²

So for 600 rotation,

➳ 30.8 × 600

➳ Area = 18,480 m²

∴ Area covered by the roller will be equal to 18,480 m²

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Answer 2

Given :

• Length = 3.5 m
• Radius = 1.4 m
• For leveling a ground it takes 600 rotations.

To find :

• Area of ground was leveled by the road roller.

According to the question,

➨ Curved surface area = 2πrh

$\tt{ = 2 \times \frac{22}{7} \times1.4 \times 3.5 }$

$\tt{ = \cancel \frac{215.6}{7} {}^{ \: \: \: 30.8} }$

$\tt{ =30.8 \: {m}^{2} }$

➨ For leveling a ground 600 rotation of the road roller is required = 30.8 m² × 600

= 18480 m²

.°. Area of ground was leveled by road roller is 18,480 m²....