Question

Two cones have their heights in the ratio 1:3 and their radii of their bases in the ratio 3:1. Find the ratio of their volumes.

Answer 1

Hint- In order to solve to this problem, we must choose formula of volume of cone which is equal to $\dfrac{1}{3}(\pi \times {{\text{r}}^2} \times {\text{h)}}$ along with the proper understanding of the figure that is being provided in the question.

Complete step by step answer:

Here in this question it is given the ratios of heights and bases radii of two different cones and we have to find the volume ratio.
We know that, the volume of cone is $\dfrac{1}{3}(\pi \times {{\text{r}}^2} \times {\text{h)}}$ ……………………………... (1)
Let the radius, height and Volume of 1st cone be ${{\text{r}}_{1{\text{ }}}}{\text{, }}{{\text{h}}_1}{\text{ and }}{{\text{v}}_1}$ respectively
And
 The radius, height and Volume of 2nd cone be ${{\text{r}}_{{\text{2 }}}}{\text{, }}{{\text{h}}_2}{\text{ and }}{{\text{v}}_2}$ respectively
So Using expression (1)
Volume of 1st cone ${{\text{v}}_1}{\text{ will be }}\dfrac{1}{3}(\pi \times {{\text{r}}_1}^2 \times {{\text{h}}_1}{\text{)}}$ …………………………………...(2)
 And volume of 2nd cone ${{\text{v}}_2}{\text{ will be }}\dfrac{1}{3}(\pi \times {{\text{r}}_2}^2 \times {{\text{h}}_2}{\text{)}}$ …………………………………….(3)
Now on dividing equation (2) by (3),
 ${\text{Hence }}\dfrac{{{{\text{v}}_1}}}{{{{\text{v}}_2}}} = \dfrac{{\dfrac{1}{3}(\pi \times {{\text{r}}_1}^2 \times {{\text{h}}_1}{\text{)}}}}{{\dfrac{1}{3}(\pi \times {{\text{r}}_2}^2 \times {{\text{h}}_2}{\text{)}}}}$

On cancelling $\dfrac{1}{3}$ and $\pi $ from numerator and denominator,
$\therefore \dfrac{{{{\text{v}}_1}}}{{{{\text{v}}_2}}} = \dfrac{{({{\text{r}}_1}^2 \times {{\text{h}}_1}{\text{)}}}}{{({{\text{r}}_2}^2 \times {{\text{h}}_2}{\text{)}}}}$
$\therefore \dfrac{{{{\text{v}}_1}}}{{{{\text{v}}_2}}} = {\left( {\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}}} \right)^2} \times \left( {\dfrac{{{{\text{h}}_1}}}{{{{\text{h}}_2}}}} \right)$ ………………………………….(4)
In the given question it is given that
 $\therefore \left( {\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}}} \right) = $ $\dfrac{3}{1}$ And $\left( {\dfrac{{{{\text{h}}_1}}}{{{{\text{h}}_2}}}} \right) = $$\dfrac{1}{3}$
Now putting the value of $\left( {\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}}} \right)$ and $\left( {\dfrac{{{{\text{h}}_1}}}{{{{\text{h}}_2}}}} \right)$ in equation (4)
$\therefore \dfrac{{{{\text{v}}_1}}}{{{{\text{v}}_2}}}$ $ = {\left( {\dfrac{3}{1}} \right)^2} \times \left( {\dfrac{1}{3}} \right)$
On solving
$\therefore \dfrac{{{{\text{v}}_1}}}{{{{\text{v}}_2}}} = $ $3$
Hence, the volume ratio of given cones will be $\dfrac{3}{1}$

Note: -Whenever we face such types of problems the key concept we have to remember is that always remember the formula of volume of cone which is stated above, then using this formula calculate the ratio of volume of two cones and simplify, we will get the required answer.