Question

A roller 1.5 metre long has a diameter of 70 cm. How many revolutions will it take to level a playground measuring 50 m × 33 m?​

Answer 1

Answer:

500

Step-by-step explanation:

lenght of roller or height = 1.5 m

diameter of roller =0.7 m

radius of roller = 0.7/2 m

CSA of cylinder = 2πrh

= 2 × 22/7 × 0.7/2 × 1.5

= 3.3 m sq

area of field = 50×33 = 1650 m sq

no. of revolution taken = 1650/3.3 = 500revolu.

Answer 2

Answer:

Number of revolution is 500

Curve surface area of cylinder = 2πrh

Step-by-step explanation:

Explanation:

Given, lenght of the roller = 1.5 m

diameter of roller = 70cm = 0.7m

so, the radius of the roller  = 0.7/2 m

we need to find  the number of revolutions will it take to level a ground

50m × 33m in dimension.

Step 1:

Curve surface area of cylinder =2πrh ..........(i)

Now , put the value of h, and r in the equation (i)

we have ,

2× 22/7 × 0.7/2 ×1.5

⇒3.3msq

Step 2:

lenght of the field = 50m

and width of the field = 33 m

so, area of the field = l×b =50 ×33 = 1650m sq.

Step 3:

No. of revolution taken by the roller = (area of field ÷CSA of cylinder)

(where CSA is curve surface area of cylinder)

= 1650/3.3 = 500 revolution.

Final answer :

Hence , the number of revolution  = 500.