Question

the volume of a metallic cylindrical pipe is 748cm³ and its length is 14 cm and its external radius is 9 cm. find its thickness.​

Answer 1

Answer:

1cm.

Step-by-step explanation:

We know that,

$volume \: of \: hollow \: cylinder = \pi( {r1}^{2} - {r}^{2} )h$

where r1 is the outer or external radius.

On substituting the given values, we have,

$volume \: of \: hollow \: cylinder = \frac{22}{7} ( {9}^{2} - {r}^{2}) \times 14$

$44( {9}^{2} - {r}^{2} )= 748$

$81 - {r}^{2} = 17$

${r}^{2} = 81 - 17$

${r}^{2} = 64$

$r = \sqrt{64}$

$r = 8cm$

Thickness = outer radius (r1) - inner radius (r)

Thickness = 9 - 8

Thickness = 1cm.

Therefore, the thickness of the hollow cylinder is 1cm.

Hope it helps you.

Good Luck

Answer 2

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

• The volume of a metallic cylindrical pipe is 748cm³ and its length is 14 cm and its external radius is 9 cm. find its thickness

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

• External Radius of pipe ( R ) = $\sf\green{ 9cm}$
• Height of the pipe ( h ) = $\sf\green{ 14cm}$
• Volume of the pipe ( V ) = $\sf\green{748 {cm}^{3} }$

━━━━━━━━━━━━━━━━━━━━━━━━━

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

$\red{\boxed{ \sf \blue{ Let \: r\: be \: the \: internal \: radius \: in \: centimetres }}}$

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

$\qquad \quad {:} \longrightarrow \sf{\sf{748 \: = \: \frac{22}{7} ( \: {9}^{2} - {r}^{2} \: ) \: \times \: 14 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{748 \: = \: \frac{22}{ \cancel{7}} ( \: {9}^{2} - {r}^{2} \: ) \: \times \: \cancel{ 14 }}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{748 \: = \: 22 \: \times \: 2 \: ( \: {9}^{2} - {r}^{2} \: ) }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{748 \: = \: 44\: ( \: {9}^{2} - {r}^{2} \: ) }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{748}{44} \: = \: 81 - {r}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\cancel \frac{748}{44} \: = \: 81 - {r}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{17 \: = \: 81 - {r}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{{r}^{2} \: = \: 81 - 17 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{{r}^{2} \: = \: 64 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{{r}\: = \: \sqrt{64} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{radius\: = \: 8cm }}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\quad {:} \longrightarrow \sf\green{\sf{ Thickness \: = \:(\: R \: - \ r\:) }}$

$\quad {:} \longrightarrow \sf\green{\sf{ Thickness \: = \: (9 \: - \ 8 ) cm}}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{ \: Thickness\: = \: 1cm }}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\bf \therefore \; Thickness= 1cm$

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More For Knowledge:-

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