A copper ball with a diameter of 5.00 cm is subjected to a pressure increase of 2.50 × 10⁶ Pa. The bulk modulus for copper is 1.40 × 10¹¹ N/m². What is the magnitude of the resulting change in the volume? A) 1.01 × 10^-3 cm³ B) 1.17 × 10^-3 cm³ C) 1.20 × 10^-3 cm³ D) 1.58 × 10^-3 cm³ E) 1.70 × 10^-3 cm³

Question

Quizplus · Accepted Answer

The change in volume ($\Delta V$) due to pressure ($P$) can be calculated using the formula $\Delta V = \frac{-\Delta P \cdot V_0}{B}$, where $V_0$ is the original volume and $B$ is the bulk modulus. The original volume of the copper ball can be found using the formula for the volume of a sphere, $V_0 = \frac{4}{3}\pi r^3$, where $r$ is the radius. Given the diameter is 5.00 cm, the radius is 2.50 cm or 0.025 m. Thus, $V_0 = \frac{4}{3}\pi (0.025)^3$ m$^3$. Substituting the given values, $P = 2.50 \times 10^{16}$ Pa and $B = 1.40 \times 10^{11}$ N/m$^2$, and converting the final volume change back to cm$^3$ (since $1 m^3 = 10^6 cm^3$), we find the change in volume to be approximately $1.17 \times 10^{-3}$ cm$^3$, which corresponds to choice B.

A Copper Ball with a Diameter of 5