A rectangular field is $70 m$ long and $60 m$ broad. A well of dimensions $14 m \times 5 m \times 6 m$ is dug outside the field and the earth dugout from this well is spread evenly in the field. How much will the earth level rise?

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Solution

Volume of dugout = Volume of the cuboid $= 14 \times 5 \times 6$
$= 70 \times 6 = 420 m^{3}$

This volume would make a cuboid on the field with 70m as length and 60m as width and let height be $x$
$\Rightarrow 70 \times 60 \times x = 420$
$\Rightarrow x = \frac{420}{60 \times 70} = \frac{420}{42 \times 100} = \frac{1}{10} m$
$\Rightarrow x = 10 c m$

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A rectangular field is 70m long and 60m broad. A well of dimensions 14m×5m×6m is dug outside the field and the earth dugout from this well is spread evenly in the field. How much will the earth level rise?

Volume of dugout = Volume of the cuboid =14×5×6=70×6=420m3 This volume would make a cuboid on the field with 70m as length and 60m as width and let height be x⇒70×60×x=420 ⇒x=42060×70=42042×100=110m⇒x=10cm

A rectangular field is $70 m$ long and $60 m$ broad. A well of dimensions $14 m \times 5 m \times 6 m$ is dug outside the field and the earth dugout from this well is spread evenly in the field. How much will the earth level rise?