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