Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 609: 97

$(B)~9.2\times 10^{-5}m^3$

We know that $\Delta V=\frac{\rho_w gh(\frac{4}{3}\pi r^3)}{B}$ $\implies \Delta V=\frac{\rho_w gh(\frac{\pi}{6}D^3)}{B}$ We plug in the known values to obtain: $\Delta V=\frac{(1027Kg/m^3)(9.81m/s^2)(923m)(\frac{\pi}{6}(1.45m)^3)}{16\times 10^{10}N/m^2}$ $\Delta V=9.2\times 10^{-5}m^3$