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:

$V_{balls}=\frac{4}{3} \pi r^{3}$ (4)

$V_{balls}=\frac{4}{3} \pi (1.5 cm)^{3}$ (5)

$V_{balls}=14.137 cm^{3}$ (6)

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

$n=\frac{V_{room}}{V_{balls}}$ (7)

$n=\frac{55,047,949.78 cm^{3}}{14.137 cm^{3}}$  (8)

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