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# Two cones have their heights in the ratio 1:3 and the radii of their bases are in the ratio 3:1, then the ratio of their volumes is:(a) 1:3(b) 27:1(c) 3:1(d) 1:27  Hint: Assume the radius and height of first cone ${{r}_{1}}$ and ${{h}_{1}}$ respectively and radius and height of second cone ${{r}_{2}}$ and ${{h}_{2}}$ respectively. Use the formula for their volumes and take their ratio to find the required ratio of their volumes.

Let us assume the height of two cones is ${{h}_{1}}$ and ${{h}_{2}}$ respectively. So, their radius can be assumed as ${{r}_{1}}$ and ${{r}_{2}}$ respectively. The volume of a cone is given as $\dfrac{1}{3}\pi {{r}^{2}}h$, that is one-third of the volume of the cylinder having the same height and radius. Therefore,

Volume of first cone $={{V}_{1}}=\dfrac{1}{3}\pi {{r}_{1}}^{2}{{h}_{1}}$

And, volume of second cone $={{V}_{2}}=\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}$

Taking the ratio of their volume we get,

$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{1}{3}\pi {{r}_{1}}^{2}{{h}_{1}}}{\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}}$

Since, $\dfrac{1}{3}\text{ and }\pi$ are constant, therefore, they get cancelled.

$\therefore \dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\times \dfrac{{{h}_{1}}}{{{h}_{2}}}$

We have been given, $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1}{3}\text{ and }\dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{3}{1}$. Substituting these values in the volume ratio we get,

$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{1}{3} \right)}^{2}}\times \dfrac{3}{1}=\dfrac{1}{9}\times \dfrac{3}{1}=\dfrac{1}{3}$

Converting this into ratio we get,

${{V}_{1}}:{{V}_{2}}=1:3$

Hence, option (a) is the correct answer.

Note: Constant term gets cancelled directly in a ratio whereas we need relation between variables to cancel them. Here, we can see that the ratio of the volume of two cones is 1:3, which is less than 1. That means that the first cone is of smaller capacity as compared to the second cone.

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