Question

If the radius of the base and height of the cone are doubled, then by how many times its volume will increase?

Answer 1

Hint: The volume of a cone having the radius ‘r’ and the height ‘h’ is given by the formula \[V = \dfrac{1}{3}\pi {r^2}h\] . In this question first we will find the volume of a normal cone and then we will find the volume of the cone whose radius and the height is doubled and then we will compare both the volume to find the increase of the volume.

Complete step-by-step answer:
Let the radius of the cone be ‘r’.
Height of the cone be ‘h’.
Hence the volume of the cone will be \[V = \dfrac{1}{3}\pi {r^2}h - - (i)\]
Now it is said that the radius of the base and height of the cone are doubled so the new radius and the height of the cone will become
 \[r' = 2r\] and \[h' = 2h\]
So the volume of the new cone whose radius and the height is doubled will be
  \[V' = \dfrac{1}{3}\pi {\left( {r'} \right)^2}\left( {h'} \right)\]
This will be equal to
 \[
\Rightarrow V' = \dfrac{1}{3}\pi {\left( {2r} \right)^2}\left( {2h} \right) \\
   = \dfrac{8}{3}\pi {r^2}h - - (ii) \\
 \]
Now we compare equation (ii) with the equation (i), hence we can say
 \[
\Rightarrow V' = 8\left( {\dfrac{1}{3}\pi {r^2}h} \right) \\
   = 8V \;
 \]
From this we can say when the radius and the height are doubled then the volume becomes 8 times the volume with radius r and height h.
So, the correct answer is “8 TIMES”.

Note: In general when we draw both the cone where height is h and radius is r with the cone of doubled height and radius we will see then figure will be much larger than the original figure so we can also say that when the radius and the height are doubled then the volume will also increase.