Question

A well with 10 m inside diameter is dug 14 m deep. The soil taken out is spread all around to a width of 5 m to form a circular embankment. Find the height of the embankment.​

Answer 1

Answer :-

$\: \\ \large{\underline{\underline{\sf{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}}$

Here the concept of Volume of Cylinder and Volume of Hollow Cylinder has been used. We know that, the volume of earth dug out from the well will be equal to the Volume of the Embankment made. Also if we will add the radius of well with the width of embankment, we can find the radius of the solid including embankment and well. Now, subtracting these radius and using the volume of Hollow Cylinder, we can find our answer.

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★ Formula Used :-

$\:\\\large{\boxed{\sf{Volume\;of\;Cylinder\;\; = \; \bf{\pi r^{2}h}}}}$

$\: \large{\sf{Radius_{(including\:well\:and\:embankment)}, \;R\: \; = \:\; \bf{Radius\;of\;Well\;+\;Width\;of\;Embankment}}}$

$\: \\\large{\boxed{\sf{Volume\;of\;Hollow\;Cylinder\;=\;\bf{\pi (R^{2}\;-\;r^{2})h}}}}$

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★ Question :-

A well with 10 m inside diameter is dug 14 m deep. The soil taken out is spread all around to a width of 5 m to form a circular embankment. Find the height of the embankment.

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★ Solution :-

Given,

» Height of well at which soil is taken out = 14 m

» Diameter of well = d = 10 m

» Radius of well = r = ½ × 10 = 5 m

» Width of the embankment = 5 m

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~ For the Volume of Soil taken out :-

$\: \\\qquad\large{\sf{:\longrightarrow\;\;\;Volume\;of\;Cylinder_{(soil\:taken\:out)}\;\; = \; \bf{\pi r^{2}h}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\;Volume\;of\;Cylinder_{(soil\:taken\:out)}\;\; = \; \bf{\dfrac{22}{\cancel{7}}\:\times\:(5)^{2}\:\times\:\cancel{14} \:\; = \;\: \underline{\underline{1100 \;\; m^{3}}}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Volume\;of\;Soil\;taken\;out\;=\;\bf{1100\;\;m^{3}}}}}}$

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~ For the Radius of solid including well and Embankment :-

$\: \\ \qquad \large{\sf{:\longrightarrow\;\:\; Radius_{(including\:well\:and\:embankment)}, \;R\: \; = \:\; \bf{Radius\;of\;Well\;+\;Width\;of\;Embankment}}}$

$\: \\ \qquad \large{\sf{:\longrightarrow\;\:\; Radius_{(including\:well\:and\:embankment)}, \;R\: \; = \:\; \bf{5\;+\;5 \;\: = \:\;\underline{\underline{10\;\:m}}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Radius\;of\;Solid_{(including\:embankment\:and\:well)}\; =\;\bf{10\;\;m}}}}}$

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~ For the Height of the Embankment :-

We know that if we will subtract the Volume of soil taken out from Volume of Whole Solid, we will get volume of embankment. And embankment is a hollow cylinder. So,

$\: \\ \qquad \large{\sf{:\Longrightarrow\;\:\;Volume\;of\;Hollow\;Cylinder_{(embankment)} \;=\;\bf{\pi (R^{2}\;-\;r^{2})h}}}$

Also we know that Volume of Soil taken out is equal to the Volume of Embankment.

Then,

$\: \\ \qquad \large{\sf{:\Longrightarrow\:\;\;1100 \;=\;\bf{\dfrac{22}{7}\:\times\:[(10)^{2}\;-\;(5)^{2})]\:\times\:h}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow\:\;\;1100 \;=\;\bf{\dfrac{22}{7}\:\times\:(100\;-\;25)\:\times\:h}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow\:\;\;1100 \;=\;\bf{\dfrac{22}{7}\:\times\:75\:\times\:h}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow\:\;\; h \;\; = \;\; \bf{\dfrac{1100\;\times\;7}{22\;\times\;75} \;\; = \; \; \underline{\underline{4.66\;\; m}}}}}$

$\: \\ \large{\underline{\underline{\rm{Height\;of\;the\;embankment\;is\;\;\boxed{\bf{4.66\;\:m}}}}}}$

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$\: \\ \large{\underbrace{\underbrace{\sf{More\;Formulas\;to\;know\;:-}}}}$

• Volume of Cube = (Side)³

• Volume of Hemisphere = ⅔ × πr³

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cone =  ⅓ × πr²h

• TSA of Hollow Cylinder = 2π(R - r)h + 2π(R² - r²)

• CSA of Hollow Cylinder = 2π(R - r)h

★ Refer to figure :-

We see the darkened part is the width of embankment and the lightened part is the well.

Answer 2

Answer:

Step-by-step explanation:

$\huge{\sf{Required \ Answer}}$

We need to find the height of the embankment. For this we need the volume of the well ie the dug soil for well, and some concepts.

Volume of Cylinder = $\sf{\pi \times r^2 \times h}$

Given, 10 m as diameter.

So, radius will be 10/2 = 5 metre.

Volume of Well = $\sf{\dfrac{22}{7} \times (25)^2 \times 14}$

Volume of Well = 1100 m³.

Now, we need to find the height of embankment.

So, here the width 5 metre should be added with radius if well, so that we can get the bigger Radius.

= 5+5

= 10 metre.

Now, volume will be the same or 1100 m³, same soil is displaced.

1100 = $\sf{\dfrac{22}{7} \times ((10)^2-(5)^2)\times h}$

h = $\sf{\dfrac{22}{7} \times 85 \times \dfrac{1}{1100}}$

Calculating, we get h as 4.66 metres (approximately).