A well of diameter 2 m is dug 14 m deep.

Solution:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

$r=\frac{2}{2}=1$

The well is $14 \mathrm{~m}$ deep. Thus, the height of the solid right circular cylinder is $h=14 \mathrm{~m}$.

 

Therefore, the volume of the solid right circular cylinder is

$V_{1}=\pi r^{2} h=\frac{22}{7} \times(1)^{2} \times 14=44$ cubic meters

Since, the embankment is to form around the right circular cylinder. Let the width of the embankment be x m. The height of the embankment is h = 40 cm = 0.4 m. Therefore, the volume of the platform is

$V_{2}=\pi\left((1+x)^{2}-1^{2}\right) \times .4$

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

$V_{1}=V_{2}$

$\pi\left((1+x)^{2}-1^{2}\right) \times .4=44$

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

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

$\Rightarrow \quad x=5$ or $x=-7$

Hence, $x=5$

Hence, width =5 m