Question 1065111
 A cylindrical tin has an internal diameter of 18cm.
 It contains water to a height of 13.2m when a heavy spherical metal ball is immersed in it, what is the new height of the water level
:
Assuming the ball diameter is the same as the cylinder diameter
The radius of the ball and the cylinder = 9 cm
:
Find the volume of the water
V = {{{pi*9^2*13.2}}}
V = 3359 cu/cm
:
Find the volume of the sphere
v = {{{4/3}}}{{{pi*9^3}}}
v = 3053.63 cu/cm
This amt of water will be displaced when the ball is submerged
therefore: 3359 + 3053.63 = 6412.62 Cu/cm is the occupied volume
:
Let h = the height of water when the ball is submerged
{{{pi*9^2*h}}} = 6412.62
254.47h = 6412.62
h = {{{6412.62/254.47}}}
h = 25.2 cm is the new height of the water