Question

The difference between outside and inside surfaces of a cylindrical metallic pipe 14 cm long is 44 cm². If the pipe is made of 88 cm³ of metal, find the outer and inner radii of the pipe.​

Answer 1

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

Let assume that,

Inner radius of metallic cylindrical pipe = r cm

Outer radius of metallic cylindrical pipe = R cm

Height, h = 14 cm.

Now, According to statement

The difference between outside and inside surfaces of a cylindrical metallic pipe 14 cm long is 44 cm²

$\rm \: 2\pi \: Rh \: - \: 2\pi \: rh \: = \: 44$

$\rm \: 2\pi \: h(R - r)= \: 44$

$\rm \: 2 \times \dfrac{22}{7} \times 14 \times (R - r)= \: 44$

$\rm \: 2 (R - r)= \: 1$

$\rm\implies \:R - r = \dfrac{1}{2} - - - (1)$

Again given that, the pipe is made of 88 cm³ of metal.

$\rm \: \pi( {R}^{2} - {r}^{2})h \: = \: 88$

$\rm \: \dfrac{22}{7}(R + r)(R - r) \times 14 = 88$

$\rm \: 44(R + r)(R - r) = 88$

$\rm \: (R + r)(R - r) = 2$

$\rm \: (R + r) \times \dfrac{1}{2} = 2$

$\rm\implies \:R + r = 4 - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2R = 4 + \dfrac{1}{2}$

$\rm \: 2R = \dfrac{8 + 1}{2}$

$\rm \: 2R = \dfrac{9}{2}$

$\rm\implies \:\rm \: R = \dfrac{9}{4}$

On substituting R in equation (1), we get

$\rm \: \dfrac{9}{4} - r = \dfrac{1}{2}$

$\rm \: r = \dfrac{9}{4} - \dfrac{1}{2}$

$\rm \: r = \dfrac{9 - 2}{4}$

$\rm\implies \:\rm \: r = \dfrac{7}{4}$

So, we have

$\rm \: Outer \: radius \: of \: metallic \: cylindrical \: pipe = \dfrac{9}{4} \: cm$

and

$\rm \:Inner \: radius \: of \: metallic \: cylindrical \: pipe = \dfrac{7}{4} \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$