In a corner of a rectangular field with dimensions $35 m \times 22 m$ , 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. [HOTS]

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Solution

$We have, Length of the fiel, l = 35 m, Width of the field, b = 22 m, Depth of the well, H = 8 m and Radius of the well, R = \frac{14}{2} = 7 m, Let the rise in the level of the field be h . Now, Volume of the earth on remaining part of the field = Volume of earth dug out \Rightarrow Area of the remaining field \times h = Volume of the well \Rightarrow (Area of the field - Area of base of the well) \times h = π R^{2} H \Rightarrow (l b - π R^{2}) \times h = π R^{2} H \Rightarrow (35 \times 22 - \frac{22}{7} \times 7 \times 7) \times h = \frac{22}{7} \times 7 \times 7 \times 8 \Rightarrow (770 - 154) \times h = 1232 \Rightarrow 616 \times h = 1232 \Rightarrow h = \frac{1232}{616} ∴ h = 2 m$

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

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In a corner of a rectangular field with dimensions 35 m×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. [HOTS]

In a corner of a rectangular field with dimensions $35 m \times 22 m$ , 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. [HOTS]