Question

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Answer 1

Given :-

The earth taken out from well is spread in its embarkment .

So Volume of well = Volume of embarkment

Radius of well (r) = 10m/2= 5m

Height of well(h) = 14m

Width of embarkment (R-r)=5m→R= 10m

To Find:-

We have to find the height of embarkment .

Solution :-

A well is in the form of cylinder so,

$\underline{\boxed{\sf\ \ Volume\ of \ Well= \pi r^2 h}}$

Find volume of well with radius : 5m and height 14m

$:\implies\sf\ Volume\ of \ well= \pi \times (5)^2\times 14\\ \\ \\ :\implies\sf \ Volume \ of\ well = \pi \times 25\times 14\\ \\ \\ :\implies\sf\ Volume\ of\ well= \underline{\boxed{\purple{\sf\ 350\pi m^3}}}----- \ eq.\ (i)$

A embarkment in the form of hollow cylinder So,

$\underline{\boxed{\sf\ Volume \ of \ embarkment = \pi h(R^2-r^2)}}$

Now Volume of embarkment with R=10m, r= 5m and Height - H

$:\implies\sf\ Volume\ of \ embarkment= \pi \times \Big[(10)^2-(5)^2\Big]\times h\\ \\ \\ :\implies\sf\ Volume \ of \ embarkment= \pi \times \Big[100-25\Big]\times h\\ \\ \\ :\implies\sf\ Volume\ of \ embarkment= \underline{\boxed{\red{\sf\ 75h\pi m^3}}}----- \ eq.\ (ii)$

Volume of Well= Volume of embarkment

Now , from equation (i) and (ii)

$:\implies\sf \ 75h\not{\pi} = 350\not{\pi} \\ \\ \\ :\implies\sf h = \cancel{\dfrac{350}{75}}\\ \\ \\ :\implies\sf Height\ of \ embarkment= \underline{\boxed{\blue{\sf\ 4.67m}}}$

Answer 2

Given :-

The earth taken out from well is spread in its embarkment .

So Volume of well = Volume of embarkment

Radius of well (r) = 10m/2= 5m

Height of well(h) = 14m

Width of embarkment (R-r)=5m→R= 10m

To Find:-

We have to find the height of embarkment .

Solution :-

A well is in the form of cylinder so,

$\underline{\boxed{\sf\ \ Volume\ of \ Well= \pi r^2 h}}$

Find volume of well with radius : 5m and height 14m

$:\implies\sf\ Volume\ of \ well= \pi \times (5)^2\times 14\\ \\ \\ :\implies\sf \ Volume \ of\ well = \pi \times 25\times 14\\ \\ \\ :\implies\sf\ Volume\ of\ well= \underline{\boxed{\purple{\sf\ 350\pi m^3}}}----- \ eq.\ (i)$

A embarkment in the form of hollow cylinder So,

$\underline{\boxed{\sf\ Volume \ of \ embarkment = \pi h(R^2-r^2)}}$

Now Volume of embarkment with R=10m, r= 5m and Height - H

$:\implies\sf\ Volume\ of \ embarkment= \pi \times \Big[(10)^2-(5)^2\Big]\times h\\ \\ \\ :\implies\sf\ Volume \ of \ embarkment= \pi \times \Big[100-25\Big]\times h\\ \\ \\ :\implies\sf\ Volume\ of \ embarkment= \underline{\boxed{\red{\sf\ 75h\pi m^3}}}----- \ eq.\ (ii)$

Volume of Well= Volume of embarkment

Now , from equation (i) and (ii)

$:\implies\sf \ 75h\not{\pi} = 350\not{\pi} \\ \\ \\ :\implies\sf h = \cancel{\dfrac{350}{75}}\\ \\ \\ :\implies\sf Height\ of \ embarkment= \underline{\boxed{\blue{\sf\ 4.67m}}}$