Question

Two cones A and B have radii in the ratio of 4:3 and their heights in the ratio 3:4. The ratio of volume of cone A to that of cone B is
(a). 4 : 3
(b). 3 : 4
(c ). 2 : 3
1 : 2

Answer 1

- Hint: Take radius and height of cone A as \[{{r}_{1}}\] and \[{{h}_{1}}\]. For cone B, \[{{r}_{2}}\] and \[{{h}_{2}}\]. Thus take the ratio of volume of cone A and cone B, then substitute the ratio of radius and height.

Complete step-by-step answer: -
It is said that there are 2 cones, cone A and cone B. Let us consider the radius of cone A as \[{{r}_{1}}\]. The height of cone A can be taken as \[{{h}_{1}}\], as shown in the figure. Similarly, let the radius of cone B be \[{{r}_{2}}\] and the height of cone B as \[{{h}_{2}}\].
It is said that the radius of both cones A and B are in the ratio 4 : 3. Thus we can write,
\[{{r}_{1}}:{{r}_{2}}=4:3\]
i.e. \[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{4}{3}-(1)\]
Now, the height of cone A and cone B are in the ratio 3 : 4 i.e. \[{{h}_{1}}:{{h}_{2}}=3:4\], this can also be written as,
\[\dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{3}{4}-(2)\]
We know that volume of cone is given as, \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
The volume of cone A will be, \[{{V}_{A}}=\dfrac{1}{3}\pi {{\left( {{r}_{1}} \right)}^{2}}{{h}_{1}}\].
The volume of cone B will be, \[{{V}_{B}}=\dfrac{1}{3}\pi {{\left( {{r}_{2}} \right)}^{2}}{{h}_{2}}\].
Thus let's take the ratio of the volume of cone A to that of cone B.
\[\therefore \dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{\dfrac{1}{3}\pi {{\left( {{r}_{1}} \right)}^{2}}{{h}_{1}}}{\dfrac{1}{3}\pi {{\left( {{r}_{2}} \right)}^{2}}{{h}_{2}}}\]
We can cancel out \[\left( \dfrac{1}{3}\pi \right)\] from both the numerator and denominator. We get,
\[\therefore \dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{\left( {{r}_{1}} \right)}^{2}}{{h}_{1}}}{{{\left( {{r}_{2}} \right)}^{2}}{{h}_{2}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\times \dfrac{{{h}_{1}}}{{{h}_{2}}}\]
Thus let us substitute values in the above expression from (1) and (2).
\[\begin{align}
  & \therefore \dfrac{{{V}_{A}}}{{{V}_{B}}}={{\left( \dfrac{4}{3} \right)}^{2}}\times \dfrac{3}{4}=\dfrac{4}{3}\times \dfrac{4}{3}\times \dfrac{3}{4} \\
 & \therefore \dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{4}{3} \\
\end{align}\]
Thus the ratio of volume of cone A to that of cone B is in the ratio 4 : 3.
\[\therefore \] Option (a) is the correct answer.

Note: When you find the relation, don’t take \[{{r}_{1}}=\dfrac{4{{r}_{2}}}{3}\] and \[{{h}_{1}}=\dfrac{3{{h}_{2}}}{4}\] and then try to substitute in equation of volume and all. You will get the answer, but it does make the answer complex. It is easier to cancel out like terms than substituting relations. You can directly substitute the values of \[\dfrac{{{r}_{1}}}{{{r}_{2}}}\] and \[\dfrac{{{h}_{1}}}{{{h}_{2}}}\] like we have alone.