SOLUTION: A heavy 9cm diameter ball is placed in an empty cylindrical tin of diameter 12cm. Enough water is poured into the can to cover the ball. If the ball is then removed, how far does t

Question 1205453: A heavy 9cm diameter ball is placed in an empty cylindrical tin of diameter 12cm. Enough water is poured into the can to cover the ball. If the ball is then removed, how far does the water level fall?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

diameter of the ball = 9
radius = diameter/2 = 9/2 = 4.5
A = volume of the ball
A = (4/3)*pi*r^3
A = (4/3)*pi*(4.5)^3
A = 121.5pi

diameter of cylinder container = 12
r = radius = diameter/2 = 12/2 = 6
The water level is h = 9 so that it reaches the top of the 9 cm diameter ball.
B = combined volume of the ball and surrounding water in the cylinder
B = pi*r^2*h
B = pi*6^2*9
B = 324pi

C = volume of water only
C = difference of volumes A and B
C = B - A
C = 324pi - 121.5pi
C = 202.5pi

Use the value of C to compute the new height of the water level after the ball is removed.
The cylinder's radius r = 6 will stay the same.
V = pi*r^2*h
202.5pi = pi*6^2*h
202.5pi = 36pi*h
h = (202.5pi)/(36pi)
h = 5.625
The water level is 5.625 cm after the ball is removed.
This assumes no water spills out.

The water line started at 9 cm and now is at 5.625 cm
This is a difference of 9 - 5.625 = 3.375 cm which represents how far the water line dropped.
This converts to the mixed number 3 & 3/8 aka since 3/8 = 0.375 exactly.

Answer in decimal form 3.375 cm
Answer as a mixed number: 3 & 3/8 cm or 3 3/8 cm

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