SSC CHSL Model Paper 5

Internal diameter of the tube = 6 cm ∴ Internal radius (r) = 3 cm
Height of the tube (/r) =10 cm
Thickness of the metal = 1 cm
∴ Outer radius (R) = Thickness of themetal + Internal radius = 1 + 3 = 4 cm
∴ Outer curved surface area
= $2\mathit{\pi }\mathit{r}\mathit{h}+\mathit{\pi }{\mathit{R}}^2+\mathit{\pi }\left({\mathit{R}}^2-{\mathit{r}}^2\right)$
= $2\mathit{\pi }\left(3\right)\left(10\right)+\mathit{\pi }{\left(4\right)}^2+\mathit{\pi }\left(16-9\right)$
$=60\mathit{\pi }+16\mathit{\pi }+7\mathit{\pi }=83\mathit{\pi }\mathrm{sq}\mathrm{cm}$