A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
The inner radius of the well is $\frac{3}{2} m$ and the height is $14 m$. Therefore, the volume of the Earth taken out of it is
$V_{1}=\pi \times\left(\frac{3}{2}\right)^{2} \times 14 \mathrm{~m}^{3}$
The inner and outer radii of the embankment are $\frac{3}{2} m$ and $4+\frac{3}{2}=\frac{11}{2} m$ respectively. Let the height of the embankment be $h$. Therefore, the volume of the embankment is
$V_{2}=\pi \times\left\{\left(\frac{11}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right\} \times h \mathrm{~m}^{3}$
Since, the volume of the well is same as the volume of the embankment; we have
$V_{1}=V_{2}$
$\Rightarrow \pi \times\left(\frac{3}{2}\right)^{2} \times 14=\pi \times\left\{\left(\frac{11}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right\} \times h$
$\Rightarrow h=\frac{9 \times 14}{112}$
$\Rightarrow h=\frac{9}{8} \mathrm{~m}$
Hence, the height of the embankment is $h=\frac{9}{8} \mathrm{~m}$
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