Question

The total surface area of hollow cylinder, which is open from both sides, is $3575{\mathrm{cm}}^2$; area of the base ring is $357.5{\mathrm{cm}}^2$ and height is $14\mathrm{cm}$. Find the thickness of the cylinder.

Answer 1

Determine the thickness of the cylinder.

Given information,

Total surface area of cylinder $=3575{\mathrm{cm}}^2$

Area of the base ring $=357.5{\mathrm{cm}}^2$

Height $=14\mathrm{cm}$

Let us consider

The inner radius $=\mathrm{r}$

Outer radius $=\mathrm{R}$

Area of the base ring

$$\begin{gathered} \mathrm{\pi }({\mathrm{R}}^2-{\mathrm{r}}^2)=357.5{\mathrm{cm}}^2 \\ {\mathrm{R}}^2-{\mathrm{r}}^2=\frac{357.5}{\mathrm{\pi }} \end{gathered}$$

Value of $\mathrm{\pi }=\frac{22}{7}=3.143$

$$\begin{gathered} {\mathrm{R}}^2-{\mathrm{r}}^2=\frac{357.5}{3.143} \\ {\mathrm{R}}^2-{\mathrm{r}}^2=113.74\mathrm{sq}.\mathrm{cm} \end{gathered}$$

$(\mathrm{R}+\mathrm{r})(\mathrm{R}–\mathrm{r})=113.74\mathrm{sq}.\mathrm{cm}\dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

Total surface area of a hollow cylinder = outer curved surface $+$ inner curved surface area $+2$(Area of the circular base)

$$\begin{gathered} 3575=2\mathrm{\pi Rh}+2\mathrm{\pi rh}+2\mathrm{\pi }({\mathrm{R}}^2-{\mathrm{r}}^2) \\ 3575=2\mathrm{\pi Rh}+2\mathrm{\pi rh}+2\times 357.5 \\ 3575=2\mathrm{\pi h}(\mathrm{R}+\mathrm{r})+715 \\ 2\mathrm{\pi h}(\mathrm{R}+\mathrm{r})=3575–715 \\ 2\mathrm{\pi h}(\mathrm{R}+\mathrm{r})=2860 \\ (\mathrm{R}+\mathrm{r})=\frac{2860}{(2\times 3.143\times 14)} \\ (\mathrm{R}+\mathrm{r})=\frac{2860}{88.004} \\ (\mathrm{R}+\mathrm{r})=32.498\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(2\right) \end{gathered}$$

Substitute equation $\left(2\right)$ in equation $\left(1\right)$ and we get

$$\begin{gathered} (32.498)\times (\mathrm{R}–\mathrm{r})=113.74 \\ (\mathrm{R}–\mathrm{r})=3.49\mathrm{cm} \end{gathered}$$

The thickness of the cylinder is $3.49\mathrm{cm}$.