Question

If the radius and the height of a right circular cone are doubled, then its volume becomes
$\begin{align}
  & \text{a) 2 times} \\
 & \text{b) 3 times} \\
 & \text{c) 4 times} \\
 & \text{d) 8 times} \\
\end{align}$

Answer 1

Hint: We know that Volume of right circular cone is given by $V=\pi {{r}^{2}}\dfrac{h}{3}$. First try to find out the volume by taking the original radius and height. Then find the volume with new radius and height which is nothing but two times the original radius and height. Substitute the values in formula to find volume and then compare the two Volumes.

Complete step-by-step answer:
Now we have a formula to find Volume of right circular cone which is $V=\pi {{r}^{2}}\dfrac{h}{3}$
Now let ${{h}_{1}}$ be the height and ${{r}_{1}}$ be the radius of the cone

Let the Volume of this original Cone be ${{V}_{1}}$.
Hence, we get the volume of original cone is ${{V}_{1}}=\pi {{r}_{1}}^{2}\dfrac{{{h}_{1}}}{3}....................(1)$
Now Consider the new cone formed.
Now the radius of new cone is two times the original radius hence new radius is $2{{r}_{1}}$ also the height is two times the original height hence new height is $2{{h}_{1}}$
Let the Volume of new cone be \[{{V}_{2}}\], hence now we have
\[\begin{align}
  & {{V}_{2}}=\pi {{(2{{r}_{1}})}^{2}}\dfrac{(2{{h}_{1}})}{3} \\
 & \Rightarrow {{V}_{2}}=4\pi {{r}_{1}}^{2}\dfrac{2{{h}_{1}}}{3} \\
 & \Rightarrow {{V}_{2}}=8\pi {{r}_{1}}^{2}{{h}_{1}} \\
\end{align}\]
Now from equation (1) we get ${{V}_{2}}=8{{V}_{1}}$
Hence, the Volume becomes 8 times the original Volume.

So, the correct answer is “Option d”.

Note: Now in the question it is given that the radius and height is doubled so it is quite tempting to select option a or option c considering the volume will be 2 times or 4 times. But the correct method to solve this question is substituting the values in formula to check the actual change in Volume.