Question

Estimate the number of Ping-Pong balls that can be packed into an average size room (without crushing them). Given that Ping-Pong ball has a radius of about 1.5 cm and assume that a typical room has dimensions12 ft ×18 ft ×9 ft.

Answer 1

Answer: 3,893,845.918 Ping-Pong balls

Explanation:

The volume of an average room is:

`V_(room)=(length)(width)(height)` (1)

`V_(room)=(12 ft)(18 ft)(9 ft)=1944 ft^(3)` (2)

Now let’s transform this
`V_(room)` to units of
`cm^(3)`, knowing
`1 ft=30.48 cm`:

`V_(room)=1944 ft^(3)\frac{{(30.48 cm)}^(3)}{1ft^(3)}=55,047,949.78 cm^(3)` (3)

On the other hand, we have Ping-Pong balls with a radius
`r=1.5 cm`, and their volume is given by:

`V_(balls)=(4)/(3) \pi r^(3)` (4)

`V_(balls)=(4)/(3) \pi (1.5 cm)^(3)` (5)

`V_(balls)=14.137 cm^(3)` (6)

Now, the number
`n` of Ping-Pong balls that can be packed into the room is:

`n=(V_(room))/(V_(balls))` (7)

`n=(55,047,949.78 cm^(3))/(14.137 cm^(3))` (8)

`n=3,893,845.918` This is the number of Ping-Pong balls that can be packed into an average size room