Question

the diameter of a road roller, 1m 40cm long is 80cm. if it takes 600 revolutions to level a playground, find the cost of leveling the ground at Rs 15 per square meter.

Answer 1

Answer:

the cost of leveling the ground at Rs 15 per square meter = RS. 31651.2

Step-by-step explanation:

It is given that,the diameter of a road roller, 1m 40cm long is 80cm.  

and it takes 600 revolutions to level a playground.

Curved surface area CSA of cylinder = 2πrh

r - radius and h height

consider the roller  as a cylinder.

To find the curved surface area of roller

r = 80/2 = 40cm = 0.4m

h = 1m 40cm = 1.4 m

C.S.A 2πrh = 2 x 3.14 x 0.4 x 1.4 =3.5168 square meter.

To find the cost

It is given that,the cost of leveling the ground at Rs 15 per square meter.

And it takes 600 revolutions to level a playground

Therefore cost = 3.5168  x 15 x 600 =31651.2

Total cost = Rs. 31651.2

Answer 2

Answer:So leveling of 2112 sq- m costs =31,680 Rs

Solution:

we know that road roller is in the shape of a cylinder.

here height = 140 cm

radius = 40 cm

As we know that when road rollers rolls on the road it's curved surface area is calculated

CSA of road roller=
$2\pi \: r \: h \\ \\ = 2 \times \frac{22}{7} \times 40 \times 140 \\ \\ = 2 \times 22 \times 40 \times 20 \\ \\ = 44 \times 800 \\ \\ =35200 \: {cm}^{2}$
So in one revolution it covers 35200 sq-cm

in 600 revolution it covers
$35200 \times 600 \\ \\ =21120000 {cm}^{2} \\ \\ = 2112 {m}^{2}$
if leveling of 1 sq-m costs 15 Rs

So leveling of 2112 sq- m costs
$= 2112 \times 15 \\ \\ =31,680 Rs$