Question

The diameter of a road roller is 90 cm and is 2m 10 cm in length. If it takes 400 revolutions to level a playground, find the cost of leveling the ground at Rs 2.25 per square meter

Answer 1

Answer:

The cost of   leveling the ground is  Rs 5340.6

Step-by-step explanation:

Given:

Diameter of the road roller =  90 cm

Length of the road roller = 2m 10 cm

Total number of number of revolution = 400

Cost of leveling = Rs 2.25 per square meter

To Find:

Total cost of leveling the ground  = ?

Solution:

Step 1 : Finding the surface area of the road roller

Converting  cm to m

90 cm  = 0.9 m

2m 10 cm  = 2 + 0.1 m = 2.1 m

Surface area of the road roller = surface are of the cylinder

= $2\pi r h$

r= radius of the road roller

r= $\frac{0.9}{2}$

r=0.45

Now surface area  is

=>$2 \times \pi \times 0.45 \times$

=>$6.28 \times 0.45 \times 2.1$

=>$2.826 \times 2.1$

=>5.934

Step 2 : Finding the area of  the ground

Total area of the ground = Surface area  of the road roller X number of revolutions  

=>$5.934 \times 400$

=> 2373.6

Step 3 : Finding the  cost of leveling

Total cost of leveling =  area of the ground  X  Cost per square meter

=> $2373.6\times 2.25$

=>5340.6