Question

If the radius of the cone decrease by $50\%$ and height increase by $20\%$. The find the percentage change in the volume.

Answer 1

Hint: First of all, you must know the formula to calculate the volume of a cone. If $r$ is the radius of the cone and $h$ is the height of the cone, the its volume $V$ is given by $V=\dfrac{1}{3}\pi {{r}^{2}}h$ . So we will first calculate the initial volume of the cone and after that, we will calculate the final volume of the cone due to the change in its radius and height. At last, a change in volume can be found out by taking the difference between the initial and final volume. Then, to get the percentage change in volume, divide this by the original volume.

Complete step-by-step solution:
Initially, we know that the radius of the cone is $r$ and its height is $h$.

If the initial volume is denoted by ${{V}_{i}}$ , then it is given by
${{V}_{i}}=\dfrac{1}{3}\pi {{r}^{2}}h\text{ }..............\text{(1)}$
Now, let us assume that the new radius of cone is $r'$ which is decreased by 50% from its initial value
$\begin{align}
  & \Rightarrow r'=r-\dfrac{50}{100}r \\
 & \Rightarrow r'=0.5r \\
\end{align}$
And its new height is $h'$ which is increased by 20% from its initial value
$\begin{align}
  & \Rightarrow h'=h+\dfrac{20}{100}r \\
 & \Rightarrow h'=1.2h \\
\end{align}$
If final volume is denoted by ${{V}_{f}}$ , then it is given by
${{V}_{f}}=\dfrac{1}{3}\pi r{{'}^{2}}h'\text{ }..............\text{(2)}$
Putting $r'=0.5r$ and $h'=1.2h$ in the above equation, we get
$\begin{align}
  & \Rightarrow {{V}_{f}}=\dfrac{1}{3}\pi {{\left( 0.5r \right)}^{2}}1.2h \\
 & \Rightarrow {{V}_{f}}=\dfrac{0.3}{3}\pi {{r}^{2}}h \\
 & \Rightarrow {{V}_{f}}=\dfrac{1}{10}\pi {{r}^{2}}h \\
\end{align}$
If the change in volume is denoted by $\Delta V$ , then it is given by
$\Delta V={{V}_{f}}-{{V}_{i}}$
Taking the value of ${{V}_{i}}$ and ${{V}_{f}}$ from the equation (1) and equation (2), we get
$\begin{align}
  & \Rightarrow \Delta V=\dfrac{1}{10}\pi {{r}^{2}}h-\dfrac{1}{3}\pi {{r}^{2}}h \\
 & \Rightarrow \Delta V=-\dfrac{7}{30}\pi {{r}^{2}}h\text{ }.................\text{(3)} \\
\end{align}$
Now the percentage change in the volume is given by
$\Delta V \%=\dfrac{\Delta V}{{{V}_{i}}}$
Taking the value of $\Delta V$ and ${{V}_{i}}$ from the equation (3) and equation (1), we get
$\begin{align}
  & \Rightarrow \Delta V\text{ }\!\!\%\!\!\text{ }=\dfrac{-\dfrac{7}{30}\pi {{r}^{2}}h}{\dfrac{1}{3}\pi {{r}^{2}}h}\times 100\% \\
 & \Rightarrow \Delta V \% =-\dfrac{7}{10}\times 100\% \\
 & \Rightarrow \Delta V \%=-70\% \\
\end{align}$
So, the percentage change in the volume comes out to be $70\%$.

Note: While calculating the percentage change, a reference must be considered. Here our reference was the initial volume and that’s why we divided the change in volume with the initial volume in order to calculate the percentage change. Percentage change may be positive or negative, negative sign tells that the final value is smaller than the reference value and the positive sign tells that the final value is greater than the reference value.