A lead ball is thrown into a cylindrical vessel with water, the radius of the base of the cylinder is 6 cm
| May 6, 2021 | education
A lead ball is thrown into a cylindrical vessel with water, the radius of the base of the cylinder is 6 cm, and the radius of the ball is 4 cm. how much the water rises in the vessel.
We denote the radius of the cylindrical vessel by R1, and the radius of the sphere by R2
Determine the volume of the ball immersed in water.
Vball = 4 * π * R2 ^ 3/3 = 4 * π * 4 ^ 3/3 = 256 * π / 3 cm3.
The volume of the displaced liquid is equal to the volume of the immersed ball.
Vв.ж. = X * π * R2 ^ 2 = X * π * 36 cm3.
Then:
256 * π / 3 = X * π * 36.
X = 256/3 * 36 = 64/27 = 2 (10/27) cm.
Answer: The liquid in the vessel will rise by 2 (10/27) cm.
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