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Sinking Ball (Posted on 2006-01-03) Difficulty: 1 of 5
A cylindrical cup is 8 cm in diameter and 20 cm tall and is filled halfway with saltwater. The density of the saltwater is 1.1 g/cm^3.
A spherical ball has a diameter of 4 cm and a mass of 40g. How much does the level of the water rise when the ball is dropped into the cup?

  Submitted by Brian Smith    
Rating: 2.0000 (1 votes)
Solution: (Hide)
First, the volume of the ball is (4/3)*pi*(2 cm)^3 = 32*pi/3 cm^3

The gives the density of the ball as equal to (mass ball)/(volume ball)
= (40g)/(32*pi/3 cm^3) = 15/(4*pi) g/cm^3 = ~1.19 g/cm³.

Barring other forces, the ball will sink and settle on the bottom of the cup. Noting that the initial water level is higher that the diameter of the ball, the entire ball will be submerged, and therefore the volume of water displaced is equal to the volume of the ball itself. This volume (4/3*pi*r³) is divided by the cup base area (pi*r²;different r) to determine how much the water level rises.

This gives (4/3*pi*2³)/(pi*4²) = (4*2³)/(3*4²) = 2/3 cm

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AnswerK Sengupta2008-03-19 11:22:05
SolutionAnsh2006-10-07 14:48:47
No Subjectalex2006-08-07 20:51:27
re(3): My thoughtsMindRod2006-01-03 22:31:17
re: My thoughtsMindRod2006-01-03 22:26:53
re(2): My thoughtsPercy2006-01-03 20:50:02
re: My thoughtsHugo2006-01-03 16:58:49
My thoughtsFletch2006-01-03 09:32:06
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