Question

A spherical ball, \[6cm\] in diameter, is melted and cast into a conical mould, the base of which is \[12cm\] in diameter. Find the height of the cone.

Answer 1

Hint: Here, we will first find the volume of a spherical ball and the volume of a cone using their respective formulas. Then we will equate both the volumes and simplify further to get the required height of the cone.

Formula Used:
We will use the following formula:
1.Volume of a Sphere is given by the formula \[V = \dfrac{4}{3}\pi {r^3}\] where \[r\] is the radius of the spherical ball.
2.Volume of a Cone is given by the formula \[V = \dfrac{1}{3}\pi {r^2}h\] where \[r\] is the radius of the cone and \[h\] is the height of the cone.

Complete step-by-step answer:
We are given a spherical ball with \[6cm\] in diameter.
Let \[{d_1}\] be the diameter of a spherical ball, so \[{d_1} = 6cm\]
We know that a radius is given by half the diameter, so we get
\[ \Rightarrow {r_1} = \dfrac{{{d_1}}}{2} = \dfrac{6}{2} = 3cm\]
So, the radius of the spherical ball is \[3cm\].
We will now find the Volume of a Spherical ball.
\[{V_1} = \dfrac{4}{3}\pi {r_1}^3\]
By substituting the radius as 3 in the above equation, we get
\[ \Rightarrow {V_1} = \dfrac{4}{3} \times {\left( 3 \right)^3}\pi {\rm{c}}{{\rm{m}}^3}\]
By simplifying the equation, we get
\[ \Rightarrow {V_1} = 4 \times {\left( 3 \right)^2}\pi {\rm{c}}{{\rm{m}}^3}\]
Applying the exponent on the terms, we get
\[ \Rightarrow {V_1} = 4 \times 9\pi {\rm{c}}{{\rm{m}}^3}\]
By multiplying the terms, we get
\[ \Rightarrow {V_1} = 36\pi {\rm{c}}{{\rm{m}}^3}\]………………………………………………………………..\[\left( 1 \right)\]
We are given a cone with 12cm in diameter and \[h\] be the height of the cone.
Let \[{d_2}\] be the diameter of a cone, so \[{d_2} = 12cm\]
We know that a radius is given by half the diameter, so we get
\[ \Rightarrow {r_2} = \dfrac{{{d_2}}}{2} = \dfrac{{12}}{2} = 6{\rm{ }}cm\]
So, the radius of the cone is \[6{\rm{ }}cm\].
Now we will find the Volume of a Cone.
\[{V_2} = \dfrac{1}{3}\pi {r_2}^2h\]
By substituting the radius, we get
\[ \Rightarrow {V_2} = \dfrac{1}{3} \times {\left( 6 \right)^2} \times h\pi \]
By applying the exponent on the terms, we get
\[ \Rightarrow {V_2} = \dfrac{1}{3} \times 36 \times h\pi \]
\[ \Rightarrow {V_2} = 12 \times h\pi \]……………………………………………………………………\[\left( 2 \right)\]
We know that the Volume of a spherical ball is equal to the Volume of a Cone.
Volume of a spherical ball \[ = \] Volume of a cone
Substituting equation \[\left( 1 \right)\] and \[\left( 2 \right)\] in the above equation, we get
\[ \Rightarrow 36\pi = 12h\pi \]
Dividing both sides by \[12\pi \], we get
\[ \Rightarrow h = \dfrac{{36\pi }}{{12\pi }}\]
\[ \Rightarrow h = 3cm\]
Therefore, the height of the cone is \[3cm\].

Note: We are using the concept of Mensuration. If an object in one shape is melted or casted or moulded into another shape, then the volume of both the things remains the same. Both the volumes are equal to one another. We are using this concept since the volume of a spherical ball is melted to form a conical mould. We should never forget to convert the diameter into radius.