A copper ball with a diameter of $5.00 \mathrm{~cm}$ is subjected to a pressure increase of $2.50 \times 106 \mathrm{~Pa}$. The bulk modulus for copper is $1.40 \times 10^{11}\mathrm{~N} / \mathrm{m}^{2}$. What is the magnitude of the resulting change in the volume?
A) $1.70 \times 10^{-3} \mathrm{~cm}^{3}$
B) $1.01 \times 10^{-3} \mathrm{~cm}^{3}$
C) $1.58 \times 10^{-3} \mathrm{~cm}^{3}$
D) $1.17 \times 10^{-3} \mathrm{~cm}^{3}$
E) $1.20 \times 10^{-3} \mathrm{~cm}^{3}$

Question

Quizplus · Accepted Answer

The change in volume $\Delta V$ can be calculated using the formula for bulk modulus $B = -\frac{\Delta P}{\Delta V / V}$, where $\Delta P$ is the change in pressure and $V$ is the initial volume. Rearranging the formula gives $\Delta V = -\frac{\Delta P \cdot V}{B}$. The initial volume $V$ of the copper ball can be calculated using the formula for the volume of a sphere, $V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the ball. Given the diameter is $5.00 \, \mathrm{cm}$, the radius $r = 2.50 \, \mathrm{cm}$. Converting the radius to meters (since the bulk modulus is given in SI units), $r = 0.025 \, \mathrm{m}$. Thus, $V = \frac{4}{3}\pi (0.025)^3 \, \mathrm{m}^3$. Plugging in the values, $\Delta V = -\frac{2.50 \times 10^6 \cdot \frac{4}{3}\pi (0.025)^3}{1.40 \times 10^{11}}$. Calculating this gives a change in volume in cubic meters, which then needs to be converted back to cubic centimeters ($1 \, \mathrm{m}^3 = 10^6 \, \mathrm{cm}^3$). The calculation results in a change in volume of approximately $1.17 \times 10^{-3} \, \mathrm{cm}^3$, making option D the correct answer.

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A Copper Ball with a Diameter Of $5.00 \mathrm{~cm}$ Is Subjected to a Pressure Increase Of