Posted by Sally on Sunday, February 27, 2011 at 12:05pm.
The area of the graduated cylinder (g) floor = pi*5*5 = 25*pi.
Th area of the beaker (b) floor is 5*5 = 25.
(Area)g/(Area)b = 25*pi/25 = pi
Since the area of the graduated cylinder is larger than that of the beaker, the volume poured into the graduated cylinder must be pi times volume of that poured into the beaker.
Let x = volume poured into the beaker, then 80-x = volume poured into the graduated cylinder, and
pi*x = 80-x
Solve for x, then x/80 = fraction.
I get x = about 19 and 80-x = about 61.
Check and solve for h for grad cyl = pi*r^2*h and
solve for h for beaker from V= 25*h; i.e.,
h = Vg/pi*25 = about 0.8
h = Vb/25 = about 0.8
You can do it more accurately. Check my work. I'm not a math man. Perhaps Reiny or Mathmate will check it for us.
Thanks! I understand what you are doing. Just one question: how did you get 80?
Also, Is the height of the liquid in the cube beaker less than or greater than 1 cm? This was a follow up question. Since, the answer is 0.8. This would be less than, correct?
The actual values I came up with were 19.316 mL and 60.684 mL (although that is far too many significant figures).
height = volume/pi*25 for graduated cylinder
height = 60.694/25*pi = 0.773 cm height.
height = volume/25 for cubic beaker = 19.316/25 =0.773 cm height.
I rounded those in my first post to 19/25 = 0.76 which is about 0.8 and
61/pi*25 = 0.78 whichis about 0.8.