A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m × 8 m × 6 m is dug outside the field and the earth dug-out from this well is spread evenly on the field. How much will the earth level rise?

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Solution

$Dimension of the well = 14 m \times 8 m \times 6 m The volume of the dug - out earth = 14 \times 8 \times 6 = 672 m^{3} Now, we will spread this dug - out earth on a field whose length, breadth and height are 70 m, 60 m and h m, respectively . Volume of the dug - out earth = length \times breadth \times height = 70 \times 60 \times h \Rightarrow 672 = 4200 \times h \Rightarrow h = \frac{672}{4200} = 0.16 m We know that 1 m = 100 cm ∴ The earth level will rise by 0.16 m = 0.16 \times 100 cm = 16 cm .$

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