A well of diameter $ 2 \mathrm{~m} $ is dug $ 14 \mathrm{~m} $ deep. The earth taken out of it is spread evenly all around it to form an embankment of height $ 40 \mathrm{~cm} $. Find the width of the embankment.

Given:

A well of diameter \( 2 \mathrm{~m} \) is dug \( 14 \mathrm{~m} \) deep. The earth taken out of it is spread evenly all around it to form an embankment of height \( 40 \mathrm{~cm} \).

To do:

We have to find the width of the embankment.

Solution:

Diameter of the well $= 2\ m$

This implies,

Radius of the well $r=\frac{2}{2}$

$=1 \mathrm{~m}$

Depth of the well $h=14 \mathrm{~m}$

Therefore,

Volume of the earth dug out $=\pi r^{2} h$

$=\frac{22}{7} \times 1 \times 1 \times 14$

$=44 \mathrm{~m}^{3}$

The height of the embankment $=40 \mathrm{~cm}$

$=\frac{40}{100}$

$=\frac{2}{5}$
Let the width of the embankment $=x \mathrm{~m}$

This implies,

Outer radius $\mathrm{R}=(1+x) \mathrm{m}$

Volume of the earth used in the embankment $=\pi(\mathrm{R}^{2}-r^{2}) \times h$

Volume of the earth dug out $=$ Volume of the earth used in the embankment

Therefore,

$\pi h(\mathrm{R}^{2}-r^{2})=44$

$\Rightarrow \frac{22}{7} \times \frac{2}{5}[(1+x)^{2}-(1)^{2}]=44$

$\Rightarrow 1+x^{2}+2 x-1=44 \times \frac{7 \times 5}{22 \times 2}$

$\Rightarrow x^2+2x=35$

$\Rightarrow x^{2}+2 x-35=0$

$\Rightarrow x^{2}+7 x-5 x-35=0$

$\Rightarrow x(x+7)-5(x+7)=0$

$\Rightarrow(x+7)(x-5)=0$

$(x+7)=0$ or $x-5=0$

If $x+7=0$, then $x=-7$ which is not possible as $x$ cannot be negative.

Therefore,

$x-5=0$

$x=5$

The width of the embankment is $5 \mathrm{~m}$.

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Updated on: 10-Oct-2022

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