A well with $10\ m$ inside diameter is dug $8.4\ m$ deep. Earth taken out of it is spread all around it to a width of $7.5\ m$ to form an embankment. Find the height of the embankment.

 Given:

A well with $10\ m$ inside diameter is dug $8.4\ m$ deep. Earth taken out of it is spread all around it to a width of $7.5\ m$ to form an embankment. 

To do:

We have to find the height of the embankment.

Solution:

Diameter of the well $= 10\ m$

This implies,

Radius $(r) =\frac{10}{2}$

$= 5\ m$

Depth of the well $(h) = 8.4\ m$

Therefore,

Volume of the earth dugout $= \pi r^2h$

$=\frac{22}{7} \times 5 \times 5 \times 8.4$

$=660 \mathrm{~m}^{3}$

Width of the embankment $=7.5 \mathrm{~m}$

This implies,

Outer radius $(\mathrm{R})=5+7.5$

$=12.5 \mathrm{~m}$

Area of the embankment $=\pi(\mathrm{R}^{2}-r^{2})$

$=\frac{22}{7}[(12.5)^{2}-(5)^{2}]$

$=\frac{22}{7}[12.5+5][12.5-5]$

$=\frac{22}{7} \times 17.5 \times 7.5$

$=22 \times 2.5 \times 7.5\ m^2$

Let $h$ be the height of the embankment.

Area of the embankment $\times h=660$

$22 \times 2.5 \times 7.5 \times h=660$

$h=\frac{660}{22 \times 2.5 \times 7.5}$

$h=1.6 \mathrm{~m}$

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Updated on: 10-Oct-2022

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