Question

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

Answer 1

$\tt\Huge{\orange{\underline{\blue{ Answer :-}}}}$

$\:\:$

$\underline\bold{\large{\gray{\textbf{Given :-}}}}$

$\:\:$

• Diameter of well = 3 m

$\:\:$

• Height of well = 14 m

$\:\:$

• Width of circular ring around the well

$\:\:$

$\underline\bold{\large{\gray{\textbf{To \: Find :-}}}}$

$\:\:$

• The height of embankment

$\:\:$

$\underline{\large{\gray{\textbf{Solution :-}}}}$

$\:\:$

We know that ,

$\:\:$

➠ $\sf Radius = \dfrac { Diameter } { 2 }$

$\:\:$

$\sf Radius = \dfrac { 3} { 2 }$

$\:\:$

Volume of the earth taken out of the well = Volume of well

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder :}}}$

$\:\:$

➠ πr²h ⚊⚊⚊⚊ ⓵

$\:\:$

Where ,

$\:\:$

• r = Radius

• h = Height

$\:\:$

$\underline{\bold{\texttt{Volume of cylindrical well :}}}$

$\:\:$

• r = $\sf \dfrac { 3 } { 2 }$

• h = 14

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 3 } { 2 } × \dfrac { 3 } { 2 } × 14$

$\:\:$

➜ 11 × 3 × 3

$\:\:$

➜ 11 × 9

$\:\:$

➜ 99 cu. m. ⚊⚊⚊⚊ ⓶

$\:\:$

• Hence the volume of well is 99 cu. m. thus the volume of earth taken out is 99 cu. m.

$\:\:$

External radius = Radius of well + Width of circular ring around well

$\:\:$

➜ External radius = $\sf \dfrac { 3 } { 2 } + 4$

$\:\:$

➜ External radius = $\sf \dfrac { 11 } { 2 }$

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder with external radius :}}}$

$\:\:$

• r = $\sf \dfrac { 11 } { 2 }$

• h = H = Height of embankment

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 11 } { 2 } × \dfrac { 11 } { 2 } × H$

$\:\:$

➜ $\sf \dfrac { 11 × 11 × 11 } { 7 × 2 } × H$

$\:\:$

➜ $\sf \dfrac { 1331 } { 14 } × H$ cu. m. ⚊⚊⚊⚊ ⓷

$\:\:$

• Hence the volume of cylinder with external radius is $\sf \frac { 1331 } { 14 } × H$ cu. m.

$\:\:$

Internal radius = Radius of well

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder with internal radius :}}}$

$\:\:$

• r = $\sf \dfrac { 3 } { 2 }$

• h = H = Height of embankment

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 3 } { 2 } × \dfrac { 3 } { 2 } × H$

$\:\:$

➜ $\sf \dfrac { 11 × 3 × 3 } { 7 × 2 } × H$

$\:\:$

➜ $\sf \dfrac { 99 } { 14 } × H$ ⚊⚊⚊⚊ ⓸

$\:\:$

• Hence the volume of cylinder with internal radius is $\sf \frac { 99 } { 14 } × H$ cu. m.

$\:\:$

Volume of cylinder with external radius - Volume of cylinder with Internal radius = Volume of embankment around the well

$\:\:$

⓷ - ⓸

$\:\:$

➜ $\sf \dfrac { 1331 } { 14 } × H - (\dfrac { 99 } { 14 } × H )$

$\:\:$

➜ $\sf \dfrac { H } { 14 } × 1331 - 99$

$\:\:$

➜ $\sf \dfrac { H } { 14 } × 1232$

$\:\:$

➜ 88 × H cu. m. ⚊⚊⚊⚊ ⓹

$\:\:$

• Hence volume of embankment is 88H cu. m.

$\:\:$

As earth taken out from well is spread to form embankment

$\:\:$

So,

$\:\:$

Volume of embankment = Volume of well

$\:\:$

⓹ = ⓶

$\:\:$

➜ 88 × H = 99

$\:\:$

➜ $\sf H = \dfrac { 99 } { 88 }$

$\:\:$

➨ H = 1.125 m

$\:\:$

• Hence the height of the embankment is 1.125 m

$\:\:$

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Answer 2

$\tt\Huge{\orange{\underline{\blue{ Answer :-}}}}$

$\:\:$

$\underline\bold{\large{\gray{\textbf{Given :-}}}}$

$\:\:$

• Diameter of well = 3 m

$\:\:$

• Height of well = 14 m

$\:\:$

• Width of circular ring around the well

$\:\:$

$\underline\bold{\large{\gray{\textbf{To \: Find :-}}}}$

$\:\:$

• The height of embankment

$\:\:$

$\underline{\large{\gray{\textbf{Solution :-}}}}$

$\:\:$

We know that ,

$\:\:$

➠ $\sf Radius = \dfrac { Diameter } { 2 }$

$\:\:$

$\sf Radius = \dfrac { 3} { 2 }$

$\:\:$

Volume of the earth taken out of the well = Volume of well

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder :}}}$

$\:\:$

➠ πr²h ⚊⚊⚊⚊ ⓵

$\:\:$

Where ,

$\:\:$

r = Radius

h = Height

$\:\:$

$\underline{\bold{\texttt{Volume of cylindrical well :}}}$

$\:\:$

r = $\sf \dfrac { 3 } { 2 }$

h = 14

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 3 } { 2 } × \dfrac { 3 } { 2 } × 14$

$\:\:$

➜ 11 × 3 × 3

$\:\:$

➜ 11 × 9

$\:\:$

➜ 99 cu. m. ⚊⚊⚊⚊ ⓶

$\:\:$

Hence the volume of well is 99 cu. m. thus the volume of earth taken out is 99 cu. m.

$\:\:$

External radius = Radius of well + Width of circular ring around well

$\:\:$

➜ External radius = $\sf \dfrac { 3 } { 2 } + 4$

$\:\:$

➜ External radius = $\sf \dfrac { 11 } { 2 }$

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder with external radius :}}}$

$\:\:$

r = $\sf \dfrac { 11 } { 2 }$

h = H = Height of embankment

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 11 } { 2 } × \dfrac { 11 } { 2 } × H$

$\:\:$

➜ $\sf \dfrac { 11 × 11 × 11 } { 7 × 2 } × H$

$\:\:$

➜ $\sf \dfrac { 1331 } { 14 } × H$ cu. m. ⚊⚊⚊⚊ ⓷

$\:\:$

• Hence the volume of cylinder with external radius is $\sf \frac { 1331 } { 14 } × H$ cu. m.

$\:\:$

Internal radius = Radius of well

$\:\:$

$\underline{\bold{\texttt{Volume of cylinder with internal radius :}}}$

$\:\:$

r = $\sf \dfrac { 3 } { 2 }$

h = H = Height of embankment

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ πr²h

$\:\:$

➜ $\sf \dfrac { 22 } { 7 } × \dfrac { 3 } { 2 } × \dfrac { 3 } { 2 } × H$

$\:\:$

➜ $\sf \dfrac { 11 × 3 × 3 } { 7 × 2 } × H$

$\:\:$

➜ $\sf \dfrac { 99 } { 14 } × H$ ⚊⚊⚊⚊ ⓸

$\:\:$

• Hence the volume of cylinder with internal radius is $\sf \frac { 99 } { 14 } × H$ cu. m.

$\:\:$

• Volume of cylinder with external radius - Volume of cylinder with Internal radius = Volume of embankment around the well

$\:\:$

⓷ - ⓸

$\:\:$

➜ $\sf \dfrac { 1331 } { 14 } × H - (\dfrac { 99 } { 14 } × H )$

$\:\:$

➜ $\sf \dfrac { H } { 14 } × 1331 - 99$

$\:\:$

➜ $\sf \dfrac { H } { 14 } × 1232$

$\:\:$

➜ 88 × H cu. m. ⚊⚊⚊⚊ ⓹

$\:\:$

Hence volume of embankment is 88H cu. m.

$\:\:$

As earth taken out from well is spread to form embankment

$\:\:$

So,

$\:\:$

Volume of embankment = Volume of well

$\:\:$

⓹ = ⓶

$\:\:$

➜ 88 × H = 99

$\:\:$

➜ $\sf H = \dfrac { 99 } { 88 }$

$\:\:$

➨ H = 1.125 m

$\:\:$

Hence the height of the embankment is 1.125 m

$\:\:$

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