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Question
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 = πR2H
⇒ (lb - πR2) × h = πR2H
`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.
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