nine solid metal spheres, each of radius 2 cm, are dropped into a cylinder partly filled with water. If the spheres are totally immersed, find the rise in the water if the radius of the cylinder is 3cm
volume of one ball = (4/3)π(2^3) = 32π/3 cm^3
so 9 of them have a volume of 288π cm^3
volume of water in cylinder = πhr^2
9πh = 288π
h = 288/9 cm or 32 cm
the water will rise 32 cm
To find the rise in the water level when the solid metal spheres are dropped into the cylinder. We can calculate the volume of water displaced by the spheres.
1. First, let's calculate the volume of each sphere using the formula for the volume of a sphere:
V = (4/3) * π * r^3
Given that the radius of each sphere is 2 cm, substituting the value into the formula:
V = (4/3) * π * (2 cm)^3 = (4/3) * π * 8 cm^3 ≈ 33.51 cm^3
2. Since there are nine spheres, we can calculate the total volume of water displaced by all the spheres:
Total Volume = 9 * 33.51 cm^3 = 301.59 cm^3
3. The rise in the water level is equal to the volume of water displaced by the spheres. This means that the rise in the water level is equal to 301.59 cm^3.
Therefore, the water level in the cylinder will rise by approximately 301.59 cm^3 when the nine solid metal spheres, each with a radius of 2 cm, are dropped into the cylinder with a radius of 3 cm.