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

The correct option is A: $3cm$

External (outer) Radius, $R=8cm$

Height, $h=10cm$

Inner radius, $r=?$

Total surface Area of hollow metal cylinder $=2\pi Rh+2\pi rh+2\pi (R^2-r^2)$

It is given that the total surface area of the given hollow cylinder is $338\pi .$

$\therefore 338\pi =2\pi (8\times 10+r\times 10+8^2-r^2)$

$\Rightarrow \frac{338\pi }{2\pi }=80+10r+64-r^2$

$\Rightarrow 169=144+10r-r^2$

$\Rightarrow 169-144+r^2-10r=0$

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

$\Rightarrow r^2-2\times r\times 5+5^2=0$

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

$\Rightarrow r-5=0or,r-5=0$

$\Rightarrow r=5or,r=5$

$\therefore r=5cm$

Thickness $=R-r=8-5=3cm$