A ball is subjected to a pressure of 25.0 atmospheres. The original diameter of the ball is 4.00 cm. The squeezed diameter of the ball is 3.99 cm. The volume strain of the ball is
Question
A ball is subjected to a pressure of 25.0 atmospheres. The original diameter of the ball is 4.00 cm. The squeezed diameter of the ball is 3.99 cm. The volume strain of the ball is
Solution
The volume strain of a material is defined as the change in volume divided by the original volume.
First, we need to calculate the original volume and the squeezed volume of the ball. We can use the formula for the volume of a sphere, which is V = 4/3 * π * r³, where r is the radius of the sphere.
The original radius of the ball is half of the original diameter, so r = 4.00 cm / 2 = 2.00 cm.
The original volume of the ball is V = 4/3 * π * (2.00 cm)³ = 33.51 cm³.
The squeezed radius of the ball is half of the squeezed diameter, so r = 3.99 cm / 2 = 1.995 cm.
The squeezed volume of the ball is V = 4/3 * π * (1.995 cm)³ = 33.49 cm³.
The change in volume is the original volume minus the squeezed volume, so ΔV = 33.51 cm³ - 33.49 cm³ = 0.02 cm³.
Finally, the volume strain is the change in volume divided by the original volume, so volume strain = ΔV / V = 0.02 cm³ / 33.51 cm³ = 0.000597, or 5.97 x 10^-4.
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