## College Physics (4th Edition)

(a) The specific gravity of the disk is $0.910$ (b) The bottom surface of the disk is $1.27~cm$ below the water level. (c) The top surface of the disk is $0.13~cm$ above the water level.
(a) We can find the density of the disk: $\rho_d = \frac{m}{V} = \frac{8.16~kg}{8.97\times 10^{-3}~m^3} = 910~kg/m^3$ We can find the specific gravity of the disk: $\frac{910~kg/m^3}{1000~kg/m^3} = 0.910$ The specific gravity of the disk is $0.910$ (b) We can find the height of the disk: $h~A = V$ $h = \frac{V}{A}$ $h = \frac{8.97\times 10^{-3}~m^3}{0.640~m^2}$ $h = 0.0140~m = 1.40~cm$ If the disk is floating in the water, then the buoyant force is equal in magnitude to the weight of the disk. According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V_w$ be the volume of the disk that is submerged. Note that $V_w$ is the volume of water that is displaced. We can find the ratio $\frac{V_w}{V}$: $\rho_w~V_w~g = \rho_d~V~g$ $\rho_w~V_w = \rho_d~V$ $\frac{V_w}{V} = \frac{\rho_d}{\rho_w}$ $\frac{V_w}{V} = \frac{910~kg/m^3}{1000~kg/m^3}$ $\frac{V_w}{V} = 0.910$ 91% of the disk is submerged. We can find the distance below the water level of the bottom surface of the disk: $(0.91)(1.40~cm) = 1.27~cm$ The bottom surface of the disk is $1.27~cm$ below the water level. (c) Since the total height of the disk is $1.40~cm$, the top surface of the disk is $0.13~cm$ above the water level.