Question

a roller 1.5 metre long has a diameter of 70 cm how many revolutions will it make to level a playground is in 50 M and 33 m

Answer 1

radius of roller (R) = (70/2) = 35 cm or 0.35 m
curved surface area of roller = 2πRH = 2(22/7)x0.35x1.5 = 3.3 sqare meter
area od ground to be plane = 50 x 33 = 1650 sq. m.
no. of revolution = 1650 / 3.3 = 500 revolutions

Answer 2

Answer:

Number of revolutions = 500 revolutions

Step-by-step explanation:

Given data

Length of the roller = 1.5 metre

diameter of roller d = 70 cm

70 cm = 0.7 m       [ ∵ 1 m = 100 cm ]

dimensions of play ground = 50 m × 33 m

here we need to how many revolutions will roller need to make to level the given playground  

Number of revolutions

= area of the play ground/ area of lateral surface of roller  

Area of the playground = 50 m × 33 m = 1650 m²

As we know that roller is in cylindrical shape

here length of the roller will be the height of the cylinder, h = 1.5 m

                                               Radius of the roller r = 0.7/2 = 0.35 m

lateral surface area of roller = 2πrh

                                              = $2(\frac{22}{7} )(0.35)(1.5)$

                                              = 2(22)(0.05)(1.5)  

                                              = 3.3 m²

Number of revolutions = 1650/ 3.3 = 500 revolutions