Question

A cylindrical well is 21 m deep it's outer amd inner diameter are 21 m and 14 m respectively find the cost of renovating the inner curved surface and outer curved surface at the rate of ₹25 per m ²

WHAT Is the ANSWER​

Answer 1

Answer:

Its outer and inner diameters are 21m & 14m respectively. Find the cost of renovating the inner curved surface and outer curved surface at the rate of ₹25 per m². And the answer is ₹57,750

Answer 2

Answer:

Given :

• A cylindrical well is 21 m deep.
• It's outer and inner diameter are 21 m and 14 m.
• The cost of renovation is ₹25 per m².

$\begin{gathered}\end{gathered}$

To Find :

• The cost of renovating the inner curved surface and outer curved surface at the rate of ₹25 per m².

$\begin{gathered}\end{gathered}$

Using Formulas :

${\longrightarrow{\small{\underline{\boxed{\sf{R = \dfrac{D}{2}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{CSA \: of \: cylinder = 2\pi rh }}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Total \: area = Area_{(Inner \: cylinder)} + Area_{(Outer \: cylinder)}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Cost = Total \: area \times Rate \: of \: renovating \: per \: {m}^{2} }}}}}}$

Where :-

• R = radius
• D = Diameter
• π 22/7
• h = Height
• CSA = Curved surface area

$\begin{gathered}\end{gathered}$

Solution :

Finding the radius of Inner cylinder :-

${\longrightarrow{\sf{Radius_{(Inner)} = \dfrac{Diameter}{2}}}}$

${\longrightarrow{\sf{Radius_{(Inner)} = \dfrac{14}{2}}}}$

${\longrightarrow{\sf{Radius_{(Inner)} = \cancel{\dfrac{14}{2}}}}}$

${\longrightarrow{\sf{Radius_{(Inner)} = 7\: m}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Radius_{(Inner)} = 7\: m}}}}}}$

∴ The radius of inner cylinder is 7 m.

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Finding the radius of Outer cylinder :-

${\longrightarrow{\sf{Radius_{(Outer)} = \dfrac{Diameter}{2}}}}$

${\longrightarrow{\sf{Radius_{(Outer)} = \dfrac{21}{2}}}}$

${\longrightarrow{\sf{Radius_{(Outer)} = \cancel{\dfrac{21}{2}}}}}$

${\longrightarrow{\sf{Radius_{(Outer)} = 10.5 \: m}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Radius_{(Outer)} = 10.5\: m}}}}}}$

∴ The radius of Outer cylinder is 10.5 m.

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Finding curved surface area of Inner cylinder :-

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 2\pi rh }}}$

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 2 \times \dfrac{22}{7} \times 7 \times 21 }}}$

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 2 \times \dfrac{22}{\cancel{7}} \times \cancel{7} \times 21 }}}$

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 2 \times 22 \times 21 }}}$

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 44\times 21 }}}$

${\longrightarrow{\sf{CSA_{(Inner \: cylinder)} = 924 \: {m}^{2}}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{CSA_{(Inner \: cylinder)} = 924 \: {m}^{2}}}}}}}$

∴ The CSA of inner circle is 924 m².

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Finding curved surface area of Outer cylinder :-

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} = 2\pi rh }}}$

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} = 2 \times \dfrac{22}{7} \times 10.5 \times 21 }}}$

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} = 2 \times \dfrac{22}{\cancel{7}} \times 10.5 \times \cancel{21} }}}$

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} = 2 \times 22\times 10.5 \times 3}}}$

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} = 44\times 31.5}}}$

${\longrightarrow{\sf{CSA_{(Outer\: cylinder)} =1386 \: {m}^{2}}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{CSA_{(Outer\: cylinder)} =1386 \: {m}^{2}}}}}}}$

∴ The CSA of outer circle is 1386 m².

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Finding total area of outer and inner cylinder :-

${\longrightarrow{\sf{Total \: area = Area_{(Inner \: cylinder)} + Area_{(Outer \: cylinder)}}}}$

${\longrightarrow{\sf{Total \: area = 924 +1386}}}$

${\longrightarrow{\sf{Total \: area = 2310 \: {m}^{2}}}}$

${\bigstar{\underline{\boxed{\sf{\red{Total \: area = 2310 \: {m}^{2}}}}}}}$

∴ The total area is 2310 m².

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Now, finding the total cost of renovating cylinder well at the rate of ₹25 m² :-

${\longrightarrow{\sf{Cost = Total \: area \times Rate \: of \: renovating \: per \: {m}^{2}}}}$

${\longrightarrow{\sf{Cost = 2310 \times 25}}}$

${\longrightarrow{\sf{Cost = Rs.57750}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Cost = Rs.57750}}}}}}$

∴ The total cost of renovating cylindrical well is Rs.57750.

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Learn More :

$\begin{gathered}\boxed{\begin{minipage}{6.2 cm}\bigstar \:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:Curved \: Surface \: Area =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:Total \: Surface \: Area = 2 \pi r(h + r)\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}}\end{gathered}$

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