A cylindrical disk has volume \(8.97 \times 10^{-3} \mathrm{m}^{3}\) and mass \(8.16 \mathrm{kg} .\) The disk is floating on the surface of some water with its flat surfaces horizontal. The area of each flat surface is $0.640 \mathrm{m}^{2} .$ (a) What is the specific gravity of the disk? (b) How far below the water level is its bottom surface? (c) How far above the water level is its top surface?

Specific gravity is the ratio of the density of a substance to the density of a reference substance, which is typically water in physics problems. To compute the specific gravity of the disk, we first need to find its density. Density (\(\rho\)) is defined as mass (\(m\)) divided by volume (\(V\)): \(\rho = \frac{m}{V}\) Given the mass of the disk (\(m = 8.16 \mathrm{kg}\)) and its volume (\(V = 8.97 \times 10^{-3} \mathrm{m}^{3}\)), we can compute the density: \(\rho = \frac{8.16}{8.97 \times 10^{-3}} \approx 909.7 \,\mathrm{kg/m^3}\) The density of water is approximately \(1000 \,\mathrm{kg/m^3}\), so the specific gravity (SG) of the disk can be computed as follows: \(\text{SG} = \frac{\rho_{\text{disk}}}{\rho_{\text{water}}} = \frac{909.7}{1000} \approx 0.91\)

Since the disk is floating on water, the buoyant force acting on it equals its weight. According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by the disk. Let \(V_{\text{displaced}}\) be the volume of water displaced by the submerged part of the disk. The weight of the displaced water is: \(W_{\text{water}} = V_{\text{displaced}} \cdot \rho_{\text{water}} \cdot g\) where \(g\) is the acceleration due to gravity. The weight of the disk is: \(W_{\text{disk}} = m_{\text{disk}} \cdot g\) Since the disk is in equilibrium, \(W_{\text{water}} = W_{\text{disk}}\). Therefore, \(V_{\text{displaced}} \cdot \rho_{\text{water}} \cdot g = m_{\text{disk}} \cdot g\) From this, we can calculate \(V_{\text{displaced}}\): \(V_{\text{displaced}} = \frac{m_{\text{disk}}}{\rho_{\text{water}}} = \frac{8.16}{1000} = 8.16 \times 10^{-3} \,\mathrm{m^3}\)

Now that we have the volume of water displaced by the disk (\(V_{\text{displaced}} = 8.16 \times 10^{-3} \mathrm{m}^{3}\)), we can determine the height of the submerged portion of the disk (\(h_{\text{submerged}}\)). Since the volume of a cylinder is equal to its cross-sectional area multiplied by its height, we can use the area of a flat surface (A) to find the height: \(h_{\text{submerged}} = \frac{V_{\text{displaced}}}{A} = \frac{8.16 \times 10^{-3}}{0.640} \approx 0.0127 \,\mathrm{m}\) The distance from the water level to the bottom surface of the disk is equal to the height of the submerged portion, so: \(h_{\text{bottom}} = h_{\text{submerged}} \approx 0.0127 \,\mathrm{m}\)

To find the distance from the water level to the top surface of the disk (\(h_{\text{top}}\)), we need to calculate the total height of the disk. Using the given volume (\(V = 8.97 \times 10^{-3} \mathrm{m}^{3}\)) and the area of a flat surface (A), we can find the total height: \(h_{\text{total}} = \frac{V}{A} = \frac{8.97 \times 10^{-3}}{0.640} \approx 0.0140 \,\mathrm{m}\) Since \(h_{\text{top}}\) is the total height minus the height of the submerged portion, we can calculate: \(h_{\text{top}} = h_{\text{total}} - h_{\text{submerged}} \approx 0.0140 - 0.0127 \approx 0.0013 \,\mathrm{m}\) So, the answers are (a) Specific gravity of the disk: \(\text{SG} \approx 0.91\), (b) Distance from the water level to the bottom surface: \(h_{\text{bottom}} \approx 0.0127 \,\mathrm{m}\), and (c) Distance from the water level to the top surface: \(h_{\text{top}} \approx 0.0013 \,\mathrm{m}\).