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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. (use =22 7 )

Vidyadip · Accepted Answer

The height of the hollow cylinder is 14cm. 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 ⇒ 2R = 5 ⇒ 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 3cm respectively.

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