A field roller, in the shape of a cylinder is used to level a rectangular field 572 m long and 250 m wide. If the radius of the roller is 7 m and its length is 65 m, how many revolutions will it take to level the field? -

50

Area of rectangular field = l × b = 572 × 250

C.S.A. of roller (cylinder)= 2 \(\pi \)rh = 2 × \(\frac{22}{7}\) × 7 × 65

Number of revolutions = \(\frac{Area\;of\;field}{C.S.A.}\) 

                                    = \(\frac{(572 × 250)}{2 × \frac{22}{7} × 7 × 65}\)

                                    = 50