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 p cm². 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

Given,

Total surface area of hollow cylinder = 338π cm²

Height = 10 cm

External radius R = 8 cm

Let internal radius = r cm

∴ 2πRh + 2πrh + 2π(R² - r²) = 338 π

= Rh + rh + (R + r) (R - r) = \(\cfrac{338\pi}{2\pi}\) = 169

=80 +10r + (8 + r) (8 - r) =169

= 10r + 64 - 8r + 8r - r² = 89

=r² - 10r + 25 = 0

= r(r - 5) -5(r - 5) =0

= r = 5 cm

∴ Thickness of hollow cylinder = (R - r) = 8 - 5 = 3 cm