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If the radii of the bases of a cylinder and a cone are in the ratio $3: 4$ and their heights are in the ratio $2: 3$, find the ratio of their volumes.
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Let radius of cylinder = $r_1$& height of cylinder = $h_1$
Let radius of cone = $r_2$& height of cone = $h_2$
$\frac{r_1}{r_2} = \frac{3}{4}$, $\frac{h_1}{h_2} = \frac{2}{3}$ (1)
$\frac{\text{volume of cylinder}}{\text{volume of cone}} = \frac{\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = 3 \times \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$ (1)
$= 3 \times \left(\frac{3}{4}\right)^2 \times \left(\frac{2}{3}\right)$
$= 3 \times \frac{9}{16} \times \frac{2}{3} = \frac{9}{8}$ (1)
Hence, required ratio is $9:8$
Let radius of cone = $r_2$& height of cone = $h_2$
$\frac{r_1}{r_2} = \frac{3}{4}$, $\frac{h_1}{h_2} = \frac{2}{3}$ (1)
$\frac{\text{volume of cylinder}}{\text{volume of cone}} = \frac{\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = 3 \times \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$ (1)
$= 3 \times \left(\frac{3}{4}\right)^2 \times \left(\frac{2}{3}\right)$
$= 3 \times \frac{9}{16} \times \frac{2}{3} = \frac{9}{8}$ (1)
Hence, required ratio is $9:8$