Question

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?

Answer 1

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

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