Question

The difference between outside and inside surfaces of a cylindrical metal pipe is 14cm long is 44 Sq.cm. If the pipe is made of 99 cubic metres of metal, find the outer and inner radius of the pipe.​

Answer 1

Let, R be the external radius and r be the inner radius of the metallic pipe. h=14cm

$\small \bold{Outer \: surface \: area−inner \: surface \: area=44 cm^2}$

$\bold{∴2πRh−2πrh=44cm^2}$

$\small { \boxed{ \boxed{ \bold \red{\therefore \: R-r = \frac{44}{2 \times \frac{22}{7} \times 14} = \frac{44 \times 7}{2 \times 22 \times 14}}}}}$

$\small { \boxed{ \boxed{ \bold \green{\therefore \: R-r = \frac{1}{2} }}}}$

$\small \bold{The \: volume \: of \: metal \: used =99cm^3}$

$\small \bold{∴ External \: volume − internal \: volume =99cm^3}$

$\bold{∴πR^2h−πr^2 h=99cm^3}$

$\bold{∴πh(R^2 −r^2 )=99cm^3}$

$\small \therefore\frac{22}{7} ×14(R+r)(R−r)=99cm^3$

[∵a^2−b^2=(a+b)(a−b)] and[∵R−r=1/2]

$22×2(R+r) \frac{1}{2} = 99 \: {cm}^{2}$

$∴R+r= \frac{99}{22} = \frac{9}{2}$

$R+r= \frac{9}{2}$

$\huge \frac{R - r = \frac{1}{2} }{ \frac{2R = \frac{10}{2} }{2R = 5} }$

$R = \frac{5}{2} = 2.5 \: cm$

∴ External radius =2.5cm

$R+r= \frac{9}{2}$

r+2.5=4.5

r=4.5−2.5=2cm

Hence Internal radius =2cm