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# A well with $10m$ inside diameter is dug $14m$ deep. Earth taken out of it is spread all around to a width of $5m$ to form an embankment; the height of the embankment is ____________$A.{\text{ 2}}{\text{.46 m}} \\ {\text{B}}{\text{. 3}}{\text{.56 m}} \\ {\text{C}}{\text{.}}{\text{ 4}}{\text{.66 m}} \\ {\text{D}}{\text{. 5}}{\text{.75 m}} \\$

Hint: In case, of the well, we have two radii one for inner circle and one for the outer circle formed by the embankment around the well and the height of the well just like cylinder being the well is open at the top and closed at the bottoms.
Here, we will Use$V = \pi {r^2}h$, and $V = \pi ({R^2} - {r^2})h$

Complete step by step solution: Diameter of the well is –
$D = 10\;m \\ \therefore r = \dfrac{d}{2} \\ \therefore r = \dfrac{{10}}{2} \\ \therefore r = 5m \\$
Depth, $h = 14m$
$V = \pi {r^2}h \\ V = \dfrac{{22}}{7} \times 5 \times 5 \times 14 \\$
Simplify the above equation –
$V = 1100{m^3}$
The spread all around the well of the width of $5m$ to form an embankment –
$R = 5 + 5 \\ R = 10m \\$
$V = \pi ({R^2} - {r^2})h \\ V = \dfrac{{22}}{7} \times ({10^2} - {5^2}) \times h \\$
Substitute, $V = 1100{m^3}$in the above equation.
$1100 = \dfrac{{22}}{7} \times (100 - 25) \times h$
$1100 \times 7 = 22 \times 75 \times h$
$h = \dfrac{{1100 \times 7}}{{22 \times 75}} \\ h = 4.66m \\$
Thus, the required solution is the height of the embankment is $h = 4.66m$
Note: Always double check the units of the given radii and the thickness of the object, convert it into the same format and then substitute the values in the standard general formula. When the thickness is not given or negligible use V= $\pi {r^2}h$ to find the volume of the cylindrical shape well or any object else find the relation among the two radii and its thickness. Also, remember all different properties to simplify the mathematical equations. Numerator goes to the denominator when it changes its side and vice-versa.