Question

The total surface area of a hollow metal cylinder, open at both ends, of external radius $8cm$ and height $10cm$ is $338\pi {cm}^2.$ Taking $r$ cm to be inner radius, write down an equation in $r$ and use it to find the thickness of the metal in the cylinder.

Answer 1

Let the outer radius be $R$ and inner radius be $r$

Total surface area $=$ Inner surface area $+$ outer surface area $+2$( area of thickness )

$=2\pi rh+2\pi Rh+2\pi (R^2-r^2)$

$\Rightarrow 338\pi =20\pi r+160\pi +128\pi -2\pi r^2$

$\Rightarrow 2\pi r^2=20\pi r+50\pi =0$

$\Rightarrow r^2-10r+25=0$

${(r-5)}^2=0$

$r=5cm$

So, thickness of the metal in cylinder $=R-r$

$=8-5$

$=3cm$

Hence, the answer is $3cm.$