Question

Three tennis balls are packaged in a box. The box is 12.1 centimetres long, 3.5 centimetres wide, and 3.5 centimetres tall. Each ball is 3.3 centimetres in diameter. What is the volume of the empty space in the box?

Answer 1

Hint: To solve this question, we first need to find out the total volume of the box. Then, we would have to find out the total volume of the three tennis balls. This can be achieved by using the standard formulae. After this, subtract the total volume of the tennis balls from that of the box, to obtain the volume of the remaining space.

Complete step by step solution:
To find the empty volume in the box, we would have to subtract the total volume of the tennis balls from the volume of the box.
Now, the dimensions of the box are as follows: \[12.1cm \times 3.5cm \times 3.5cm\]
Thus, the volume of the box is calculated to be \[12.1 \times 3.5 \times 3.5 = 148.225\;c{m^3}\] , by using the \[l \times b \times h\] formula for a cuboid.
Now, the diameter of the tennis ball is \[3.3cm\] .
By using the formula for the volume of a sphere, i.e., \[\dfrac{4}{3}\pi {r^3}\] , we obtain the volume of one tennis ball as follows.
  \[\dfrac{4}{3}\pi {\left( {{{3.3}}{2}} \right)^3} = {\text{18}}{\text{.816}}\;c{m^3}\]
To obtain the total volume of the tennis balls, we need to multiply the volume of one tennis ball by \[3\] .
On multiplying, we obtain: \[{\text{18}}{\text{.816}}\;c{m^3} \times 3 \approx 56.450\;c{m^3}\]
Now, we will find the remaining volume in the box by subtracting this volume by the total volume of the box. On subtraction, we will obtain the answer to be: \[148.225\;c{m^3} - 56.45\;c{m^3} = 91.775\;c{m^3}\] .
Thus, the empty space that will remain in the box after packaging will be \[91.775\;c{m^3}\] .
So, the correct answer is “$91.775\;c{m^3}$”.

Note: The volume of a sphere is the capacity that sphere has. The shape of a sphere is round and three -dimensional. It has three axes, namely x-axis, y-axis and z-axis which defines its shape. The volume depends on the diameter or radius of the sphere, because if we take the cross-section of the sphere, it is a circle.