In a cylindrical beaker with a diameter of 2.5 cm, filled with water to a certain level, 4 equal metal balls

| February 1, 2021 | education

In a cylindrical beaker with a diameter of 2.5 cm, filled with water to a certain level, 4 equal metal balls with a diameter of 1 cm are lowered. How much will the water level in the beaker change?

Let us designate the initial water level as h cm, and the water level after immersion of the balls is equal to H cm.

Let’s determine the volume of the ball immersed in water.

Vball = 4 * π * (D / 2) ^ 3/3 = 4 * π * (1/8) / 3 = π / 6 cm3.

Then the volume of the four balls is: V = 4 * Vball = 2 * π / 3 cm3.

The volume of the displaced liquid is equal to the volume of the submerged balls.

Vв.ж. = (H – h) * π * (D / 2) ^ 2 = (H – h) * π * 6.25 / 4 cm3.

Then:

2 * π / 3 = (H – h) * π * 6.25 / 4.

(H – h) = 8 / 18.75 = 32/75 cm.

Answer: The level will change by 32/75 cm.



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