Question

Please help me out. :)
Seven steel balls, each of diameter 4cm are dropped into a tall cylinder flask of radius 5cm, which contain water. By how much does the water level rise? (Assume the balls are entirely submerged)

Answer 1

Formula for the volume of a sphere: \[V = \frac{4}{3}\pi r^3\]Remember, you have the diameter of the sphere, so you'll need to account for that difference.

Compute the volume of the 7 balls. Then compute the area of the bottom of the cylinder flask. The rise in water level will be the volume of the balls divided by the area of the bottom of the cylinder. You can think of this as

\[V_{rise} = \pi r_{cyl}^2 h_{rise}\]and solving for \(h_{rise}\) as we know the other values.

Answer 2

the volume of the 7 balls is 234.51032164, right?

Answer 3

Yes, approximately :-)

Answer 4

I get 234.572

Answer 5

I would keep the problem in exact form:

volume of balls = \[7*\frac{32\pi}{3} = \frac{224\pi}{3}\]Then the area of the base is \[\pi r^2 = \pi 5^2 = 25\pi\]and the height of the water rise is \[\frac{224\pi}{3}*\frac{1}{25\pi} = \frac{224\cancel{\pi}}{3}*\frac{1}{25\cancel{\pi}} = \frac{224}{75} \approx 2.99\text { cm}\]

Answer 6

ahhh.. i got it. :) thank you. ^_^