Problem:
A sphere is dropped in a can partially filled with water. What is the rise in the height of water if they have equal diameters of 6in?
Solution:
Diameter of sphere = 6 in
r of sphere = 6/2
r of sphere = 3 in
Volume of sphere
[tex]\[\begin{array}{l}V = \frac{4}{3}\pi {r^3}\\\\V = \frac{4}{3}\pi {(3)^3}\end{array}\][/tex]
[tex]\[V = \frac{4}{3}\pi (27)\][/tex]
V = 36π
A can is a cylindrical vessel
diameter of can = 6 in
radius of can = 6/2
radius of can = 3 in
Volume of can = πr²h
Volume of can = π(3)²h
Volume of can = π(9)h
to get the rise in height, equate the volume of both sphere and cylinder
Volume of sphere = Volume of can
36π = π(9)(h)
h = 4 in
Answer:
There is a rise of 4 in height after the sphere was dropped in a can
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