The radius of a road roller is 7 m and its length is 14 m. It takes 100 revolutions to level the ground of a field. Find the area of the field. [Use π = $$\frac{22}7$$​] 

Correct option is D

Given:

Radius of the road roller = 7 m 

Length of the road roller = 14 m

The roller makes 100 revolutions 

Formula Used: 

​The circumference of a circle is given by the formula:

Circumference = 2πr 

Solution: 

$$\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \, \text{m}$$ 

​The area covered by the roller in one revolution is the area of the rectangular path formed by the length and the circumference of the roller.

The area is given by:

Area of path in one revolution = Circumference × Length 

Area of path in one revolution = 44 × 14 = 616 $$m^2$$ 

Since the roller makes 100 revolutions, the total area covered is:

Total area covered = 616 × 100 = 61600 $$m^2$$ 

Thus the area of the field is 61600$$m^2$$​.