The difference between the outer and inner curved surface areas o…

Question

The difference between the outer and inner curved surface areas of a hollow right circular cylinder $14\ cm$ long is $88\ cm^2$. If the volume of metal used in making the cylinder is $176\ cm^3$, find the outer and inner diameters of the cylinder. $\Big(\text{use}\ \pi=\frac{22}{7}\Big)$

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Answer

The height of the hollow cylinder is $14\ cm$.
Let the inner and outer radii of the hollow cylinder are r cm and R cm respectively.
The difference between the outer and inner surface area of the hollow cylinder is
$=2\pi\text{R}\times14-2\pi\text{r}\times14$
$=28\pi(\text{R}-\text{r})\text{cm}^2$ By the given condition, this difference is $88$ square cm.
Hence, we have $=28\pi\Big(\text{R}-\text{r}\Big)=88$
$\Rightarrow\text{R}-\text{r}=\frac{44\times7}{14\times22}$
$\Rightarrow\text{R}-\text{r}=\frac{44\times7}{14\times22}$
$\Rightarrow\text{R}-\text{r}=1$ The volume of the metal used in making the cylinder is
$\text{V}_1=\pi\{(\text{R})^2-(\text{r})^2\}\times14\text{cm}^3$ By the given condition, the volume of the metal is $176$ cubic cm.
Hence, we have $\pi\{(\text{R})^2-(\text{r}^2)\}\times14=176$
$\Rightarrow\text{R}^2-\text{r}^2=\frac{176\times7}{14\times22}$
$\Rightarrow\text{R}^2-\text{r}^2=4$
$\Rightarrow(\text{R}-\text{r})(\text{R}+\text{r})=4$
$\Rightarrow1\times(\text{R}+\text{r})= 4$
$\Rightarrow\text{R}+\text{r}=4$
Hence, we have two equations with unknowns $R$ and $r R - r = 1 R + r = 4$ Adding the two equations,
we have $(R - r) + (R + r) = 1 + 4 \Rightarrow 2R = 5 \Rightarrow R = 2.5$
Then from the second equation, we have $r = 4 - 2.5 = 1.5$
Therefore, the outer and inner diameters of the hollow cylinder are 5cm and $3\ cm$ respectively.