Question

The external diameter of a cylinder of a cylindrical shaped iron pipe is 25cm and its length is 20cm. If the thickness of the pipe is 1cm. Find the TSA of the pipe.

Answer 1

Dear student
$$\begin{gathered} \mathrm{External}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{pipe}\left(\mathrm{R}\right)=12.5\mathrm{cm} \\ \mathrm{height}\mathrm{of}\mathrm{the}\mathrm{pipe}\left(\mathrm{h}\right)=20\mathrm{cm} \\ \mathrm{thickness}\mathrm{of}\mathrm{the}\mathrm{pipe}\left(\mathrm{w}\right)=1\mathrm{cm} \\ \mathrm{To}\mathrm{find}\mathrm{total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{internal}\mathrm{radius}\left(\mathrm{r}\right) \\ \mathrm{W}=\mathrm{R}-\mathrm{r} \\ \Rightarrow 1=12.5-\mathrm{r} \\ \Rightarrow \mathrm{r}=12.5-1 \\ \Rightarrow \mathrm{r}=11.5\mathrm{cm} \\ \mathrm{Total}\mathrm{surface}\mathrm{area}=2\mathrm{\pi }(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r}+\mathrm{h}) \\ =2\mathrm{\pi }(12.5+11.5)(12.5-11.5+20) \\ =2\left(\frac{22}{7}\right)\left(24\right)\left(21\right) \\ =2\left(22\right)\left(24\right)\left(3\right) \\ =3168{\mathrm{cm}}^2 \end{gathered}$$
Regards