Math, asked by juhi1269, 3 months ago

The volume of metal used in making a hollow cylindrical pipe is 748cm3.its length is 14 cm and its external radius is 9 cm. find its internal radius.

Answers

Answered by SarcasticL0ve
51

\sf Given \begin{cases} & \sf{Volume\:of\:hollow\:cylinder\:pipe = \bf{748\:cm^3}}  \\ & \sf{Length\:of\: cylindrical\:pipe = \bf{14\:cm}} \\ & \sf{External\:radius\:of\: cylindrical\:pipe = \bf{9\:cm}} \end{cases}\\ \\

To find: Internal radius of pipe?

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☯ Let internal and external radius of cylindrical pipe be r and R cm respectively.

⠀⠀⠀⠀

\bf{\dag}\;{\underline{\frak{Volume\:of\: cylindrical\: pipe\:is\:given\:by,}}}\\ \\

\star\;{\boxed{\sf{\pink{Volume_{\:(cylindrical\:pipe)} = \pi (R^2 - r^2) \times h}}}}\\ \\

:\implies\sf \dfrac{22}{7} \times \bigg(9^2 - r^2 \bigg) \times 14 = 748\\ \\

:\implies\sf \dfrac{22}{ \cancel{7}} \times \bigg(81 - r^2 \bigg) \times \cancel{14} = 748\\ \\

:\implies\sf 22 \times \bigg(81 - r^2 \bigg) \times 2 = 748\\ \\

:\implies\sf 44 \times \bigg(81 - r^2 \bigg) = 748\\ \\

:\implies\sf 81 - r^2 = \cancel{\dfrac{748}{44}}\\ \\

:\implies\sf 81 - r^2 = 17\\ \\

:\implies\sf r^2 = 81 - 17\\ \\

:\implies\sf r^2 = 64\\ \\

:\implies\sf \sqrt{r^2} = \sqrt{64}\qquad\qquad\bigg\lgroup\bf Taking\:sqrt.\:both\:sides \bigg\rgroup\\ \\

:\implies{\underline{\boxed{\frak{\purple{r = 8\:cm}}}}}\;\bigstar\\ \\

\therefore\:{\underline{\sf{Internal\:radius\:of\:pipe\:is\: {\textsf{\textbf{8\:cm}}}.}}}

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\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}}

\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}}


Cosmique: Awesome! :3
Answered by Anonymous
28

{\bold{\sf{\underline{Understanding \: the \: question}}}}

❉ This question says that there is a metal used in making a hollow cylindrical pipe and it's volume is 748 cm³ Now this question says that it's length is 14 cm and it's external radius is 9 cm. Now it says that we have to find the interior radius of that object.

{\bold{\sf{\underline{Given \: that}}}}

❉ Volume of hollow cylindrical pipe = 748 cm³

❉ Length is 14 cm

❉ Exterior radius is 9 cm

{\bold{\sf{\underline{To \: find}}}}

❉ Interior radius of hollow cylindrical pipe

{\bold{\sf{\underline{Solution}}}}

❉ Interior radius of hollow cylindrical pipe = 8 cm

{\bold{\sf{\underline{Assumptions}}}}

❉ Let exterior radius is E

❉ Let interior radius is I

{\bold{\sf{\underline{Using \: concept}}}}

❉ Volume of given hollow cylindrical pipe formula ( atq )

{\bold{\sf{\underline{Using \: formula}}}}

❉ Volume of given hollow cylindrical pipe = π(E² - I²) × h

{\bold{\sf{\underline{Full \: solution}}}}

↦ Volume = π(E² - I²) × h

↦ 748 = 22/7 (9² - I²) × 14

↦ 748 = 22 (81 - I²) × 2

↦ 748 = 44 (81 - l²)

↦ 748 / 44 = 81 - l²

↦ 474 / 22 = 81 - l²

↦ 237 / 11 = 81 - I²

↦ 17 = 81 - l²

↦ 81 - 17 = l²

↦ 64 = l²

↦ √64 = l

↦ 8 = l

↦ I = 8 cm

  • Henceforth, 8 cm is interior radius of hollow cylindrical pipe.

More knowledge -

Diagram of Cylinder :

\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{r}}\put(9,17.5){\sf{h}}\end{picture}

Formulas related to Cylinder :

\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}

Formulas related to SA & Volume :

\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}


Cosmique: Perfect! :3
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