A well of inner diameter 14 m is dug to a depth of 12 m.

Solution:

Inner diameter $=14 \mathrm{~m}$

i.e., radius $=7 \mathrm{~m}$

Depth $=12 \mathrm{~m}$

Volume of the earth dug out $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 7 \times 7 \times 12=1848 \mathrm{~m}^{3}$

Width of embankment $=7 \mathrm{~m}$

Now, total radius $=7+7=14 m$

Volume of the embankment $=$ total volume $-$ inner volume

$=\pi \mathrm{r}_{\mathrm{o}}{ }^{2} \mathrm{~h}-\pi \mathrm{r}_{\mathrm{i}}{ }^{2} \mathrm{~h}=\pi \mathrm{h}\left(\mathrm{r}_{\mathrm{o}}{ }^{2}-\mathrm{r}_{\mathrm{i}}{ }^{2}\right)$

$=\frac{22}{7} \mathrm{~h}\left(14^{2}-7^{2}\right)=\frac{22}{7} \mathrm{~h}(196-49)$

$=\frac{22}{7} \mathrm{~h} \times 147=21 \times 22 \mathrm{~h}$

$=462 \times \mathrm{h} \mathrm{m}^{3}$

Since volume of embankment $=$ volume of earth dug out, we have:

$1848=462 h$

$\Rightarrow h=\frac{1848}{462}=4 m$

$\therefore$ Height of the embankment $=4 \mathrm{~m}$