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
$\pi$sq. cm . Taking r to be inner radius, find the thickness of the metal in the cylinder.

Please answer as early as possible .

Answer 1

inner radius is 5cm and thickness of metal is 3cm.

Answer 2

Answer:

Thickness of the metal in the cylinder= 3 cm

Step-by-step explanation:

The total surface area of a hollow metal cylinder open at both ends of external radius 8 cm and height 10 cm is 338 π sq. cm . Taking r to be inner radius, find the thickness of the metal in the cylinder.

As cylinder is open from both ends,so

Total surface area= Curved surface area

CSA of hollow cylinder

$2\pi \: Rh+2\pi \: rh+ 2\pi( {R}^{2} - {r}^{2} ) \\ \\ 2\pi \: 8 \times 10 +2\pi \: r \times 10 +2\pi( {8}^{2} - {r}^{2} ) = 338\pi \\ \\ 80 +10r+ (64 - {r}^{2} ) = 169 \\ \\ 10r+64 - {r}^{2} = 169 - 80 \\ \\10r+ 64 - {r}^{2} = 89 \\ \\ 10r- {r}^{2} = 64 - 89 \\ \\ {r}^{2} -10r+25 = 0 \\ \\ {r}^{2} -5r-5r+25=0\\\\({r-5})^{2}=0\\\\ r = 5$

Thickness of the metal= Outer radius-Inner Radius

= 8-5

= 3 cm