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 40 cm high, how high is the water after the sphere is placed in it?

h = 53,3 cm is the new level of water in the cylinder

Step-by-step explanation:

V₁ = volume of the sphere V₁ = (4/3) * π*r³

V₂ = volume of the cylinder V₂ = π*r²*h

Then if the sphere is totally inside water, the quantity of water sphere shifts is equal to its volume. That water will go up.

That shifts of water will occupy a new volume in the cylinder increasing the height of water according to the equation of the volume of the cylinder

That is;

(4/3) * π*r³ = π*r²*h h = the differece in height between rhe water level in the cilynder without the sphere and with the sphere

Then:

(4/3) * r = h

h = 1.33 * 10 h = 13,3 cm