Question

A solid is in the shape of a cone on a hemisphere with both their radii being equal to 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid. \[\left[ \text{Use }\pi =\dfrac{22}{7} \right]\]

Answer 1

Hint: We have given a solid which consists of a cone topped on a hemisphere in such a way that the radii of both cone and hemisphere are the same. Now, we have to find the volume of cones and the volume of hemisphere and add them to get the total volume of solid. We know that the volume of the cone is equal to $\dfrac{1}{3}\pi {{r}^{2}}h$ . In this formula, r and h are the radius and height of the cone. And the volume of the hemisphere is equal to $\dfrac{2}{3}\pi {{r}^{3}}$ where r is the radius of the hemisphere.

Complete step by step answer:
In the below figure, we have shown the solid which is given in the above question as the cone topped on the hemisphere in such a way that the radii of both the cone and the hemisphere is the same.

In the above figure, we have shown the radius of the cone and hemisphere as “r” and height of the cone as “h”.
We have given the radii of both the cone and hemisphere as 7 cm and the height of the cone is equal to the diameter of the cone.
We know that diameter is equal to double of radius so radius is given as 7 cm so diameter is given as:
$\begin{align}
  & 2\times 7cm \\
 & =14cm \\
\end{align}$
Now, we got the height of the cone as 14 cm.
To find the volume of the solid we need the volume of the cone and hemisphere and then add both of them.
The formula for volume of the cone is equal to:
$\dfrac{1}{3}\pi {{r}^{2}}h$
Substituting the value of r as 7 cm and h as 14 cm and $\pi =\dfrac{22}{7}$ we get,
$\dfrac{1}{3}\left( \dfrac{22}{7} \right){{\left( 7 \right)}^{2}}\left( 14 \right)$
One of the 7 will be cancelled out from numerator and denominator we get,
$\begin{align}
  & \dfrac{1}{3}\left( 22 \right)\left( 7 \right)\left( 14 \right) \\
 & =\dfrac{2156}{3}c{{m}^{3}} \\
\end{align}$
We have got the volume of the cone as $\dfrac{2156}{3}c{{m}^{3}}$.
We know the formula for volume of the hemisphere as:
$\dfrac{2}{3}\pi {{r}^{3}}$
Substituting the value of r as 7 cm and $\pi =\dfrac{22}{7}$ we get,
$\dfrac{2}{3}\left( \dfrac{22}{7} \right){{\left( 7 \right)}^{3}}$
One of the 7 will be cancelled out from numerator and denominator we get,
$\dfrac{2}{3}\left( \dfrac{22}{1} \right){{\left( 7 \right)}^{2}}$
$=\dfrac{2156}{3}c{{m}^{3}}$
Hence, we got the volume of the hemisphere as $\dfrac{2156}{3}c{{m}^{3}}$.
Now, adding the volume of the cone and the hemisphere we get,
$\begin{align}
  & \dfrac{2156}{3}+\dfrac{2156}{3} \\
 & =\dfrac{4312}{3}c{{m}^{3}} \\
\end{align}$
Converting the above fraction into decimal we get,
$1437.33c{{m}^{3}}$

Hence, we got the volume of the solid as $1437.33c{{m}^{3}}$.

Note: The point to be noted that the question has no particular requirement of the volume of the cone to be in certain units but sometimes in the question it demands that volume of the cone in ${{m}^{3}}\text{ or }c{{m}^{3}}\text{ }or\text{ }L$ then we have to convert our answer if it is not in the certain required units otherwise we will cost our marks in exam.
For example, if the question has asked that the volume of the solid should be in ${{m}^{3}}$ then we have to multiply the volume that we got above by ${{10}^{-6}}$ so the volume in ${{m}^{3}}$ is equal to:
$\begin{align}
  & 1437.33{{\left( 10 \right)}^{-6}}{{m}^{3}} \\
 & =0.00143733{{m}^{3}} \\
\end{align}$