Question

The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}=4620{\mathrm{cm}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{ring}=115.5{\mathrm{cm}}^2 \\ \mathrm{Height},h=7\mathrm{cm} \\ \mathrm{Let}R\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{outer}\mathrm{ring}\mathrm{and}r\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{inner}\mathrm{ring}. \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{ring}=\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ 115.5=\mathrm{\pi }\left(R^2-\mathit{}{r}^2\right) \\ {\mathrm{R}}^2-{\mathrm{r}}^2=115.5\times \frac{7}{22} \\ \mathit{(}R\mathit{+}r\mathit{)}\mathit{(}R\mathit{-}r\mathit{)}=36.75...........\left(\mathrm{i}\right) \\ \mathrm{Total}\mathrm{surface}\mathrm{area}=\mathrm{Inner}\mathrm{curved}\mathrm{surface}\mathrm{area}+\mathrm{Outer}\mathrm{curved}\mathrm{surface}\mathrm{area}+\mathrm{Area}\mathrm{of}\mathrm{bottom}\mathrm{and}\mathrm{top}\mathrm{rings} \\ 4620=2\mathrm{\pi }rh+2\mathrm{\pi }Rh+2\times 115.5 \\ 2\mathrm{\pi }h(R+r)=4620-231 \\ R+r=\frac{4389\times 7}{2\times 22\times 7} \\ R+r=\frac{399}{4}...........\left(\mathrm{ii}\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}R+r\mathrm{from}\mathrm{the}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{in}\left(\mathrm{i}\right): \\ \frac{399}{4}\mathit{(}R\mathit{-}r\mathit{)}=36.75 \\ (R-r)=36.75\times \frac{4}{399}=0.368\mathrm{cm} \\ \therefore \mathrm{Thickness}\mathrm{of}\mathrm{the}\mathrm{cylinder}=(R-r)=0.368\mathrm{cm} \end{gathered}$$