Question

What will be the volume of the following figure?

• A. $\frac{40}{3}\pi {\text{mm}}^3$
• B. $24\pi {\text{mm}}^3$
• C. $\frac{32}{3}\pi {\text{mm}}^3$
• D. $40\pi {\text{mm}}^3$

Answer 1

The correct option is A $\frac{40}{3}\pi {\text{mm}}^3$

Observe the given figure, it consists of a hemisphere and a cone.

Thus, the volume of this figure will be calculated as,

$V=\text{Volume of cone}+\text{Volume of hemisphere}$

The cone in the given figure can be drawn as,


Here, the base radius of cone is $2\text{mm}$ with centre at $O$ and $6\text{mm}$ is the depth of the cone. Therefore, the volume of cone is calculated as,

$\begin{array}{cc}\text{Volume of cone}& =\frac{1}{3}\pi r^{2}h\\ & =\frac{1}{3}\pi \left(2{)}^2\right(6)\\ & =\frac{24}{3}\pi \\ & =8\pi {\text{mm}}^3\end{array}$

Now, the hemisphere in the given figure is represented as,

The radius of hemisphere is $2\text{mm}$ with centre at $O$. Therefore, the volume of hemisphere is calculated as,

$\begin{array}{cc}\text{Volume of hemisphere}& =\frac{1}{2}\text{Volume of sphere}\\ & =\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)\\ & =\frac{2}{3}\pi r^3\\ & =\frac{2}{3}\pi (2{)}^3\\ & =\frac{16}{3}\pi {\text{mm}}^3\end{array}$

Thus, substitute the value of volume of cone and hemisphere in the equation for $V$, to calculate the volume of the given figure.

$\begin{array}{cc}V& =8\pi +\frac{16}{3}\pi \\ & =\frac{24+16}{3}\pi \\ & =\frac{40}{3}\pi \end{array}$

Hence, the volume of the given figure is $\frac{40}{3}\pi {\text{mm}}^3$.