Question

the diameter of a road roller is 140 cm and the length is 2.2 m .it takes 400 revolutions to a level a stretch 2.2 m wide. find the surface are of stretch of road leveled in meter sq.

Answer 1

given diameter of roadroller = 140cm = 1.4m
radius will be 0.7
length of roller = 2.2 m

therefore, CSA = 2×22/7×0.7×2.2
= 9.68 cm^2
It takes 400 revolutions to level a stretch 2.2m wide
therefore the surface area of road levelled =
400 × 9.68 m^2
= 3872.00 m^2

Answer 2

The surface area of stretch of road leveled is 3872 m².

Given:

• Diameter of a road roller is 140 cm.
• The length( h) is 2.2 m .
• it takes 400 revolutions to a level a stretch 2.2 m wide.

To find : find the surface are of stretch of road leveled in meter sq.

Solution:

Formula to used:

• CSA of cylinder: 2πrh unit²

Concept to be used:

• 1 revolution of road roller= CSA of cylinder

Step 1:

Find CSA of road roller.

Radius(r) =140/2= 70 cm

r= 0.70 m

Height/length(h)=2.2 m

$CSA = 2 \times \frac{22}{7} \times 2.2 \times 0.70$

or

$CSA = 2 \times22 \times 2.2 \times 0.10$

or

$\bf CSA = 9.68 \: {m}^{2}$

Step 2:

Multiply CSA with 400.

CSA multiplication with number of revolution gives the surface are of stretch of road leveled.

Area of road leveled $= 9.68 \times 400$

or

Area of road leveled$\bf = 3872 \: {m}^{2}$

Thus,

Area of road leveled is 3872 m².

Note: Because road is as wide as length of road roller, so it will complete entire width of road in one go.

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