Question

Can you solve this?

In order to promote sports in a village, Gram Panchayat of that village allocated some barren land to make a playground. In order to utilise the land for playing, a cylinderical roller of diameter 7m and length 14m was used to level it. If it takes 1500 revolutions to level the playground (rolling once), find :
a) the area of playground in hectares and
b) the values exhibited by the gram panchayat

Please solve this showing full steps.

Answer 1

the curved surface area of the cylinder multiplied by the number of times is equal to the area of the playground .

2 ˣ pi ˣ r ˣ h ˣ no: of times.

1 m² = 0.0001 hectare. so the answer is 46.2 hectares . the value is that they are very kind 

Answer 2

The cylinder tolls on the ground on the curved surface part. So each revolution of the roller covers one CSA. If 1500 revolutions leveled the playground, it means that the area of the playground is equal to 1500  * CSA of the roller.

Radius of roller = R = 7m/2 = 3.5 m
Length/height = L = 14 m
CSA = 2πR L = 2*22/7 * 3.5 * 14 = 308 m²

Area of playground:  1500 * 308 = 462, 000 m²
           1 Hectare = 10, 000 m²
So the area = 46.20 Hectares.

b) The value that is exhibited is social welfare. It is to encourage the youth for extra curricular activities. Sports facilities will help the young to spend their time more appropriately. They will do exercise their body. They also will enhance their sporting skills. All this shows that the panchayat cares about promoting the interests of their youth. Leveling the ground for the children also means that panchayat spends money for the social welfare.

     This also lays a ground for the future sportspersons from that region.