| If the ratio of the volume of two cones is 1: 6 and the ratio of the radii of their bases is 1: 2, then the ratio of their height will be |
A) 2 : 3
B) 3 : 4
C) 1: 3
D) 4 : 9
Correct Answer: A
Solution :
| Let the volume of the cones are\[{{V}_{1}}\] and \[{{V}_{2}},\] also' their radius are \[{{R}_{1}}\]and \[{{R}_{2}},\] respectively. |
| Then, according to the question, \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{1}{6}\]and \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\] |
| Let the height of the cones are \[{{H}_{1}}\] and \[{{H}_{2}}\] which are in ratio \[k:1\] |
| \[\therefore \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{k}{1}\] |
| Then, \[\frac{\frac{1}{3}\pi R_{1}^{2}{{H}_{1}}}{\frac{1}{3}\pi R_{2}^{2}{{H}_{2}}}=\frac{1}{6}\]\[\Rightarrow \]\[{{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{2}}\times \left( \frac{{{H}_{1}}}{{{H}_{2}}} \right)=\frac{1}{6}\] |
| \[\Rightarrow \]\[{{\left( \frac{1}{2} \right)}^{2}}\times \frac{k}{1}=\frac{1}{6}\]\[\Rightarrow \]\[k=\frac{2\times 2}{6}=\frac{2}{3}\] |
| \[\Rightarrow \]\[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{2/3}{1}=\frac{2}{3}\] |
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