A hollow cylinder having external radius $8cm$ and height $10cm$ has total surface area of $338\pi c{m}^2$. Find the thickness of the hollow metallic cylinder.

The correct option is B: $3cm$

As per given picture,

External radius $=R=8$ cm

Internal radius $=r$ (assume)

Height $=10$ cm

Total surface area $=$ $338\pi c{m}^2$

Now, Total surface area $=$ Area of external cylinder (curved surface area)

$+$ Area of internal cylinder (curved surface area) $+$ Twice Area of base (ring

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

$\therefore 2\pi (Rh+rh+R^2-r^2)=338\pi$

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

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

$\Rightarrow 80+10r+64-r^2=169$

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

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

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

$\Rightarrow (r-5{)}^2=0$ [Using identity, $a^2-2ab+b^2=(a-b{)}^2$]

$\Rightarrow (r-5)(r-5)=0$

$\therefore r=5$

So, thickness of metal $=R-r=(8-5)cm$ $=3cm$