SOLUTION: A heavy metal sphere with radius 10 cm is dropped into a right circular cylinder with base radius of 10 cm. If the original cylinder has water in it that is 20 cm high, how high

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A heavy metal sphere with radius 10 cm is dropped into a right circular cylinder with base radius of 10 cm. If the original cylinder has water in it that is 20 cm high, how high      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 193537This question is from textbook
: A heavy metal sphere with radius 10 cm is dropped into a
right circular cylinder with base radius of 10 cm. If the
original cylinder has water in it that is 20 cm high, how
high is the water after the sphere is placed in it?
I do not understand how to solve this. Please help. Thank you. This question is from textbook

Found 2 solutions by vleith, RAY100:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
here is a URL for volume --> http://math.about.com/od/formulas/ss/surfaceareavol.htm
Let's think about this first.
We have a sphere that will sink in the water.
The cylinder is wide enough to contain the sphere. (one could argue that a sphere with radius 10 won't fit into a cylinder with radius 10. But we will let that one go for now)
When we drop the sphere into the cylinder of water, it sinks to the bottom.
How much water is displaced by the sphere we dropped in?
The volume of a sphere is given by V+=+4%2Api%2Ar%5E3%2F3
So we know how much water is displaced.
Now that we know the amount of water that 'moved', and we know the container holding that water is a right circular cylinder, all we need to know is the formula for the volume of such a cylinder.
V+=+pi%2Ar%5E2+%2A+h+
Now we can equate those two volumes and solve for h
4+%2Api+%2A+r%5E3+%2F+3+=+pi+%2A+r%5E2+%2A+h+
4+%2A+r+%2F+3+=+h+
We are told r = 10 , so that yields h+=+40%2F3 = 13.33 centimeters

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
Volume of sphere =4/3 *pi*r^3=(4/3)pi(10)^3
Volume of cylinder = Area of Base *Height=pi(r)^2*h
the differential height in the cylinder as the sphere displaces the liquid is
found by setting sphere volume equal to cylinder volume, and solve for h
(4/3 )pi (r^3)=pi(r^2)(h)
4/3*10^3=10^2 *h
4/3 *10=h
40/3=13.33=h
Final Height = original + displaced=20 +13.33=33.33 ANSWER
,
,
,
For FUN
lets assume that the water cannot pass around sphere,but air can, after all both radius are =10, and it might be a tight fit, then what happens,
The sphere settles into the water until the water reaches the major diameter.
how much water goes into this cusp is the difference of the hemisphere volume and the cylinder volume.,.and the corresponding equivalent height of the cylinder is how far the sphere sinks into the water.
calculating
pi*10^2(10)-(4/6)pi(10^3)=pi (10^2)h
10-(4/6)10=h
3.33=h
original less 3.33= 20-3.33=16.67 as new height to bottom of sphere
height to major dia of sphere is 16.67+10=26.67, also new height of water
just for fun