Question

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly
all around it in the shape of a circular ring of width 4 m to form an embankment. Find the
height of the embankment

Answer 1

Solution:

Given:

=> Diameter of well = 3 cm

=> Radius of well = $\sf{\dfrac{3}{2}\;cm}$

=> Height (h) = 14 cm

To Find:

=> Height of embankment.

Formula used:

$\sf{\implies Volume\;of\;circle=\pi r^{2}h}$

$\sf{\implies Area\;of\;embankment = Outer\;area-Inner\;area}$

$\sf{\implies \pi R^{2}-\pi r^{2}}$

So,

$\sf{\implies Volume\;of\;the\;earth\;taken\;out\;of\;the\;well=\pi r^{2}h}$

$\sf{\implies \dfrac{22}{7}\times \dfrac{3}{2}\times \dfrac{3}{2}\times 14}$

$\sf{\implies 99\;m^{3}}$

$\sf{\implies Outer\;radius\;of\;embankment=R=\Bigg(\dfrac{3}{2}+4\Bigg)\;m=\dfrac{11}{2}\;m}$

$\sf{\implies Area\;of\;embankment = Outer\;area-Inner\;area}$

$\sf{\implies \pi R^{2}-\pi r^{2}}$

$\sf{\implies \dfrac{22}{7}\times \Bigg[\bigg(\dfrac{11}{2}\bigg)^{2}-\bigg(\dfrac{3}{2}\bigg)^{2}\Bigg]}$

$\sf{\implies \dfrac{22}{7}\times \Bigg[\dfrac{121}{4}-\dfrac{9}{4}\Bigg]}$

$\sf{\implies \dfrac{22}{7} \times \dfrac{112}{4}}$

$\sf{\implies 88\;m^{2}}$

Hence,

$\sf{\implies Height\;of\;embankment=\dfrac{Volume}{Area}}$

$\sf{\implies \dfrac{99}{88}}$

$\large{\boxed{\boxed{\red{\sf{\implies Height\;of\;embankment=1.125\;m}}}}}$

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Height=1.125\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment.

• We have to find the height of the embankment.

$\underline \bold{Given : } \\ \implies Diameter \: of \: well = 3 \: m \\ \\ \implies Depth = 14 \: m \\ \\ \implies Width \: of \: embankment = 4 \: m \\ \\ \underline \bold{To \: Find : } \\ \implies Height \: of \: embankment = ?$

• According to given question :

$\bold{By \: Volume \: of \: cylinder : } \\ \implies Volume \: of \:well = \pi {r}^{2} h \\ \\ \implies Volume = \pi ( {1.5})^{2} \times 14 \\ \\ \bold{\implies Volume = 31.5\pi m^{3}} \\ \\ \bold{For \: Outer \: radius :} \\ \implies R = ( \frac{3}{2} + 4) = \frac{11}{2} \\ \bold{Area \: of \: embankment =(Outer \: Area - Inner \: Area)} \\ \implies Area = \pi(( \frac{11}{2})^{2} - ( { \frac{3}{2} })^{2} ) \\ \\ \implies Area = \pi(\frac{121}{4} - \frac{9}{4}) \\ \\ \implies Area = \pi (\frac{121 - 9}{4} )\\ \\ \implies Area = 28\pi {m}^{2} \\ \\ \bold{ For \: height \: of \: embankment :}\\ \implies H= \frac{Volume}{Area} \\ \\ \implies H = \frac{ \cancel{31.5\pi}}{ \cancel{28\pi}} \\ \\ \bold{\implies H = 1.125 \: m} \\ \\ \bold {\therefore Height \: of \: embankment \: is \: 1.125 \: m}$