Question
The difference between outside and inside surface areas of cylindrical metallic pipe $14\ cm$ long is $44m^2$. If the pipe is made of $99cm^3$ of metal, find the outer and inner radii of the pipe.
✓
Answer
Let inner radius of pipe be $r_1$
Radius of outer cylinder be $r_2$
Length of cylinder $(h) = 14\ cm.$
Surface area of hollow cylinder $(r_2 - r_1)$ $=2\pi$
Given surface area of cylinder $= 44m^2$
$2\pi\text{Rh}-2\pi\text{r}\text{h}=44$
$2\pi\text{h}(\text{R}-\text{r})=44$
$2\times\frac{22}{7}\times14(\text{R}-\text{r})=44$
$\text{R}-\text{r}=\frac{44\times7}{2\times22\times14}$
$\text{R}-\text{r}=\frac{1}{2}\ ...(1)$
Now,
Volume of pipe $= 99cm^3$
$\pi\text{R}^2\text{h}-\pi_2\text{r}^2\text{h}=99$
$\pi\text{h}(\text{R}^2-\text{r}^2)=99$
$(\text{R}^2-\text{r}^2)=\frac{99}{\pi\times\text{h}}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99\times7}{22\times14}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99}{22\times2}$
Putting value of $\text{R}-\text{r}=\frac{1}{2}$
$(\text{R}+\text{r})\frac{1}{2}=\frac{99}{22\times2}$
$(\text{R}+\text{r})=\frac{9}{2}\ ...(2)$
Adding equation (1) and (2)
$\text{R}-\text{r}=\frac{1}{2}$
$\text{R}+\text{r}=\frac{9}{2}$
$2\text{R}=5$
$\text{R}=2.5$
And $\text{r}=\frac{9}{2}-\frac{5}{2}$
$\text{r}=2\text{cm}$
Radius of outer cylinder be $r_2$
Length of cylinder $(h) = 14\ cm.$
Surface area of hollow cylinder $(r_2 - r_1)$ $=2\pi$
Given surface area of cylinder $= 44m^2$
$2\pi\text{Rh}-2\pi\text{r}\text{h}=44$
$2\pi\text{h}(\text{R}-\text{r})=44$
$2\times\frac{22}{7}\times14(\text{R}-\text{r})=44$
$\text{R}-\text{r}=\frac{44\times7}{2\times22\times14}$
$\text{R}-\text{r}=\frac{1}{2}\ ...(1)$
Now,
Volume of pipe $= 99cm^3$
$\pi\text{R}^2\text{h}-\pi_2\text{r}^2\text{h}=99$
$\pi\text{h}(\text{R}^2-\text{r}^2)=99$
$(\text{R}^2-\text{r}^2)=\frac{99}{\pi\times\text{h}}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99\times7}{22\times14}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99}{22\times2}$
Putting value of $\text{R}-\text{r}=\frac{1}{2}$
$(\text{R}+\text{r})\frac{1}{2}=\frac{99}{22\times2}$
$(\text{R}+\text{r})=\frac{9}{2}\ ...(2)$
Adding equation (1) and (2)
$\text{R}-\text{r}=\frac{1}{2}$
$\text{R}+\text{r}=\frac{9}{2}$
$2\text{R}=5$
$\text{R}=2.5$
And $\text{r}=\frac{9}{2}-\frac{5}{2}$
$\text{r}=2\text{cm}$
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