A well of diameter $ 3 \mathrm{~m} $ is dug $ 14 \mathrm{~m} $ deep. The earth taken out of it has been spread evenly all around it to a width of $ 4 \mathrm{~m} $ to form an embankment. Find the height of the embankment.

Given:

A well of diameter \( 3 \mathrm{~m} \) is dug \( 14 \mathrm{~m} \) deep. The earth taken out of it has been spread evenly all around it to a width of \( 4 \mathrm{~m} \) to form an embankment.

To do:

We have to find the height of the embankment.

Solution:

Diameter of the well $=3\ m$

This implies,

Inner radius of the well $r=\frac{3}{2}=1.5 \mathrm{~m}$

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

This implies,

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

$=\frac{22}{7} \times 1.5 \times 1.5 \times 14$

$=99 \mathrm{~m}^{3}$
Width of the embankment $w=4 \mathrm{~m}$

This implies,

Outer radius $\mathrm{R}=1.5+4$

$=5.5 \mathrm{~m}$
Let $h$ be the height of the embankment.

Therefore,

$\pi(\mathrm{R}^{2}-r^{2}) h=99$

$\Rightarrow \frac{22}{7}[(5.5)^{2}-(1.5)^{2}] h=99$

$\Rightarrow \frac{22}{7}[30.25-2.25] h=99$

$\Rightarrow \frac{22}{7} \times 28 h=99$

$\Rightarrow h=\frac{99 \times 7}{22 \times 28}$

$\Rightarrow h=\frac{9 \times 1}{2\times4}$

$\Rightarrow h=\frac{9}{8}\ m$

The height of the embankment is $\frac{9}{8} \mathrm{~m}$. 

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Updated on: 10-Oct-2022

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