Question

in a village 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 find the hight of the embankment.​

Answer 1

${\pmb{\sf{\underline{Understanding \: the \: Question...}}}}$

★ This question says that in a village a well with 10 metres inside diameter is dug 14 metres deep earth taken out of it is spread all around to a width of 5 metres to form an embankment, we have to find the height of the embankment.

★ Diagram of a cylinder is given below, giving diagram of cylinder because we'll is in the shape of cylinder.

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${\pmb{\sf{\underline{Some \: formulas...}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Radius \: of \: cylinder \: = \:\sqrt \dfrac{v}{\pi h}}}$

${\pmb{\sf{\underline{Given \: that...}}}}$

★ A well is with 10 m inside diameter,

${\small{\underline{\boxed{\sf{Radius \: = \dfrac{Diameter}{2}}}}}} \\ \\ :\implies \sf Radius \: = \dfrac{Diameter}{2} \\ \\ :\implies \sf Radius \: = \dfrac{10}{2} \\ \\ :\implies \sf Radius \: = 5 \: metres$

★ Well is dug 14 m deep earth

★ Taken out of it is spread all around to a width of 5 m to form an embankment

${\pmb{\sf{\underline{To \: find...}}}}$

★ Height of the embankment

${\pmb{\sf{\underline{Solution...}}}}$

★ Height of the embankment = 4.66 m

${\pmb{\sf{\underline{Using \; concepts...}}}}$

★ Formula to find out the volume of cylinder =

${\small{\underline{\boxed{\sf{Volume \: = \pi r^{2}h}}}}}$

★ Formula to find out the volume of two cylinders(according to the question) from this we get the height of the embankment =

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

${\pmb{\sf{\underline{Full \; Solution...}}}}$

~ Firstly we have to use the formula to find out the volume of cylinder.

${\small{\underline{\boxed{\sf{Volume \: = \pi r^{2}h}}}}} \\ \\ :\implies \sf Volume \: = \pi r^{2}h \\ \\ :\implies \sf Volume \: = \dfrac{22}{7} \times 5 \times 5 \times 14 \\ \\ :\implies \sf Volume \: = 22 \times 5 \times 5 \times 2 \\ \\ :\implies \sf Volume \: = 22 \times 25 \times 2 \\ \\ :\implies \sf Volume \: = 44 \times 25 \\ \\ :\implies \sf Volume \: = 1100 \: metres^3$

~ As it's given that it is taken out of it is spread all around to a width of 5 m to form an embankment. Henceforth,

$:\implies \sf Radius \: = 5 + 5 \: = 10 \: m \\ \\ :\implies \sf And \: other \: radius \: is \: 5 \: m$

~ Now we have to use the formula to find out the volume of two cylinders(according to the question) from this we get the height of the embankment =

${\small{\underline{\boxed{\sf{Volume \: = \pi(R^2 - r^2)h}}}}} \\ \\ :\implies \sf Volume \: = \pi(R^2 - r^2)h \\ \\ :\implies \sf 1100 \: = \dfrac{22}{7} (10^2 - 5^2)h \\ \\ :\implies \sf 1100 \: = \dfrac{22}{7} (100-25)h \\ \\ :\implies \sf 1100 \: = \dfrac{22}{7} (75)h \\ \\ :\implies \sf 350 \: = (75)h \\ \\ :\implies \sf \dfrac{350}{75} \: = h \\ \\ :\implies \sf 4.66 \: = h \\ \\ :\implies \sf Height \: = 4.66 \: m$

4.66 m is the height.

Answer 2

Given :

• Diameter of the well = 10m
• Radius of the well (r) = 10/2 = 5m
• Height of the well (h) = 14m
• Width of an embankment = 5m

To Find :

• Height of the embankment ( H )

Solution :

Internal radius ( r ) = 5m

External radius ( R ) = 5 + 5 = 10m

★ Volume of the embankment = π ( R² - r² ) H

★ Volume of the Cylinder = πr²h

We've, R = 10, r = 5, h = 14

⇒ Volume of the embankment = Volume of the well

⇒ π ( R² - r² ) H = πr²h

⇒ ( 10² - 5² ) H = 5² × 14

⇒ ( 100 - 25 ) H = 25 × 14

⇒ 75H = 350

⇒ H = 350/75

⇒ H = 4.66

⇒ H = 4.67 ( approx )

•°• Hence, The Height of the embankment is 4.67