In a corner of a rectangular field with dimensions 35m × 22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

#### Solution

We have,

Length of the field, l = 35 m,

Width of the field, b = 22 m,

Depth of the well, H = 8 m and

Radius of the well, `"R" = 14/7 = 7 "m"`

Let the rise in the level of the field be h.

Now,

Volume of the earth on remaining part of the field = Volume of earth dug out

⇒ Area of the earth on remaining part of the field = Volume of earth

⇒ (Area of the field - Area of base of the well ) × h = πR^{2}H

⇒ (lb - πR^{2}) × h = πR^{2}H

`rArr (35xx22-22/7xx7xx7)xx"h" =22/7xx7xx7xx8 `

⇒ (770 - 154) × h = 1232

⇒ 616 × h = 1232

`⇒ "h" = 1232/616`

∴ h = 2 m

So, the rise in the level of the field is 2 m.