Question

A hollow cylindrical pipe is 21 dm long. its outer and inner diameter are 10 cm and 6 cm respectively. find the volume of the copper used in making the pipe.​

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• A hollow cylindrical pipe is 21 dm long. its outer and inner diameter are 10 cm and 6 cm respectively. find the volume of the copper used in making the pipe.

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

• Height of the Cyclindrical Pipe = 21dm
• External Diameter = 10cm
• Internal Diameter = 6cm

$\large\underline{ \underline{ \sf \maltese{ \: </strong><strong>To </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>Find</strong><strong>:- }}}$

• The Volume of the Copper used in making the Pipe.

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

$\underbrace\red{\boxed{ \sf \blue{ 1dm \: = \: 10cm }}}$

Height of the Cyclindrical Pipe = 21dm

$\qquad \quad {:} \longrightarrow \sf{\sf{21 \: \times \: 10cm }}$

$\quad {:} \longrightarrow \sf{\bf{Height \: of \: the \: Cyclindrical \: Pipe\: = \: 210cm }}$

$\underbrace\red{\boxed{ \sf \blue{ \frac{Diameter}{2} \: = \: radius }}}$

External Diameter = 10cm

$\qquad \quad {:} \longrightarrow \sf{\sf{\frac{10}{2} \: = \: radius }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\cancel\frac{10}{2} \: = \: radius }}$

$\quad {:} \longrightarrow \sf{\bf{External \: Radius \: = \: 5cm }}$

Internal Diameter = 6cm

$\qquad \quad {:} \longrightarrow \sf{\sf{\frac{6}{2} \: = \: radius }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\cancel\frac{6}{2} \: = \: radius }}$

$\quad {:} \longrightarrow \sf{\bf{Internal \: Radius \: = \: 6cm }}$

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$\underbrace\red{\boxed{ \sf \blue{ Volume \: of \: Copper\: used \: in \: making \: Pipe }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Volume \: of \: Copper \: = \: \pi \: ( {(R)}^{2} \: - \: {(r)}^{2} \: ) \: \times \: Height }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: of \: Copper \: = \: \frac{22}{7} \: ( {(5)}^{2} \: - \: {(3)}^{2} \: ) \: \times \: 210 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: of \: Copper \: = \: \frac{22}{7} \: ( {(5)}^{2} \: - \: {(3)}^{2} \: ) \: \times \: 210 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: of \: Copper \: = \: \frac{22}{\cancel7} \: ( 25 \: - \: 9 \: ) \: \times \: \cancel{210 } }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: of \: Copper \: = \: 22 \: \times( 25 \: - \: 9) \: \: \times \: 30 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: of \: Copper \: = \: 22 \: \times \: 16 \: \: \times \: 30 }}$

$\qquad \quad {:} \longrightarrow \sf{\bf{Volume \: of \: Copper \: = \: 10560{cm}^{3} }}$

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$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Volume \: of \: Copper\: used \: in \: making \: pipe \: = \underline {\underline{ 10560{cm}^{3}}}}\\\end{gathered}\end{gathered}$

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