Question

Some plastic balls of radius 1cm were melted and cast into a tube. The thickness, length and upper radius of the tube were 2cm, 9cm, and 30 cm respectively. How many balls were melted to make the tube?

Answer 1

Hint:
Here, we have to find the number of balls that were melted to make the tube with the given parameters. First, we will find the volume of the sphere with the given radius. Then we will find the volume of the cylinder with the given thickness, length and upper radius. We will then equate both the volume to find the number of balls and solve it further to get the required answer. Volume is defined as the quantity of the substance in a closed surface.

Formula Used:
We will use the following formula:
1) Volume of a sphere is given by the formula \[{V_S} = \dfrac{4}{3}\pi {r^3}\], where \[r\] is the radius of the sphere.
2) Volume of a cylinder is given by the formula \[{V_C} = \pi {r^2}h\], where \[r\] is the radius and \[h\] is the height of the cylinders.

Complete step by step solution:
We are given plastic balls of radius 1 cm.
Now, we have to find the volume of the plastic ball. We know that the plastic ball is in the form of a sphere.
Now using the formula of sphere, we get
Volume of a plastic ball\[ = \dfrac{4}{3}\pi {r^3}\]
Substituting \[r = 1\] in the above equation, we get
\[ \Rightarrow \]Volume of a plastic ball \[ = \dfrac{4}{3}\pi {\left( 1 \right)^3}\]
\[ \Rightarrow \]Volume of a plastic ball \[ = \dfrac{4}{3}\pi \]
Now, to find the volume for the number of balls the volume of a plastic ball has to be multiplied by the number of balls.
Let \[x\] be the number of balls that are melted.
Volume of \[x\]plastic balls \[ = x \times \dfrac{4}{3}\pi \]
\[ \Rightarrow \] The volume of \[x\]plastic balls \[ = \dfrac{4}{3}\pi x\]
The thickness, length and upper radius of the tube were 2cm, 9cm, and 30 cm respectively.
We know that the tube is in the form of a cylinder.
Now, we have to find the inner radius by subtracting thickness from the upper radius.
Inner Radius\[ = \] Upper Radius\[ - \]Thickness
Substituting 30 for upper radius and 2 for thickness, we get
\[ \Rightarrow \] Inner Radius\[ = 30 - 2\]
\[ \Rightarrow \] Inner Radius\[ = 28{\rm{cm}}\]
We know that length and height are the same in the case of cylinders.
Now, we have to find the volume of the tube.
We know that the volume of a cylinder is given by the formula \[{V_C} = \pi {r^2}h\].
Now as tube is a hollow cylinder, so we can write
Volume of a tube\[ = \pi {r_1}^2h - \pi {r_2}^2h\]
Substituting \[{r_1} = 30{\rm{cm}},{r_2} = 28{\rm{cm}}\] and \[h = 9{\rm{cm}}\], we get
\[ \Rightarrow \]Volume of a tube\[ = \pi {\left( {30} \right)^2} \cdot 9 - \pi {\left( {28} \right)^2} \cdot 9{\rm{c}}{{\rm{m}}^3}\]
Simplifying the expression, we get
\[ \Rightarrow \]Volume of a tube\[ = 9\pi \left( {900 - 784} \right){\rm{c}}{{\rm{m}}^3}\]
Subtracting the terms, we get
\[ \Rightarrow \]Volume of a tube\[ = 9\pi \left( {116} \right){\rm{c}}{{\rm{m}}^3}\]
Multiplying the terms, we get
\[ \Rightarrow \]Volume of a tube\[ = 1044\pi {\rm{c}}{{\rm{m}}^3}\]
We know that the Volume of \[x\] plastics balls is equal to the volume of a tube.
Volume of \[x\]plastic balls\[ = \] Volume of a tube
Substituting the values of the volumes in the above equation, we get
\[ \Rightarrow \dfrac{4}{3}\pi x = 1044\pi \]
On cross multiplication, we get
\[ \Rightarrow x = 1044 \times \dfrac{3}{4}\]
Dividing 1044 by 4, we get
\[ \Rightarrow x = 261 \times 3\]
Multiplying the terms, we get
\[ \Rightarrow x = 783\]

Therefore, 783 balls are melted to make the tube.

Note:
Here, we have used the concept of mensuration to find the number of balls. If an object in one shape is melted or casted 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 number of plastic balls are melted to form a tube. We might make a mistake by not multiplying the number of balls with the volume of a plastic ball. This will give us the wrong answer.