| If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ratio of their heights will be |
A) 3 : 8
B) 8 : 3
C) 3 : 4
D) 4 : 3
Correct Answer: B
Solution :
| \[{{V}_{1}}\,:\,{{V}_{2\,}}\,=\,2\,:\,3\]\[{{r}_{1}}:{{r}_{2}}=1:2,\]\[{{h}_{1}}:{{h}_{2}}=?\] |
| Now, \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{2}{3}\Rightarrow \frac{\frac{1}{3}\pi r_{1}^{2}{{h}_{1}}}{\frac{1}{3}\pi r_{2}^{2}{{h}_{2}}}=\frac{2}{3}\] |
| \[\Rightarrow \] \[{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\times \frac{{{h}_{1}}}{{{h}_{2}}}=\frac{2}{3}\] |
| \[\Rightarrow \] \[\frac{1}{4}\times \frac{{{h}_{1}}}{{{h}_{2}}}=\frac{2}{3}\] |
| \[\Rightarrow \] \[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{8}{3}\] |
| \[\therefore \] \[{{h}_{1}}:{{h}_{2}}=8:3\] |
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