Question

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

Answer 1

Total surface area of a hollow metal cylinder = 338 $\pi c{m}^2$

Let R be the outer radius, r be inner radius and h be the height of the cylinder.

So, $2\pi Rh+2\pi rh+2\pi R^2-2\pi r^2=338\pi$

$\Rightarrow 2\pi h(R+r)+2\pi (R^2-r^2)=338\pi$

$\Rightarrow h(R+r)+(R^2-r^2)=169$

$\Rightarrow 10(8+r)+(8^2-r^2)=169$

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

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

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

$\therefore$ Thickness of the metal = R - r = 8 - 5 = 3 cm