## College Physics (4th Edition)

The correct answer is: D. $\frac{1}{5}$
Let $V$ be the total volume of the ball. Let $m_w$ be the mass of water displaced. Let $m_b$ be the mass of the ball. Let $V_w$ be the volume of water that is displaced by the ball. We can find the ratio $\frac{V_w}{V}$: $m_w~g = m_b~g$ $m_w = m_b$ $\rho_w~V_w = \rho_b~V$ $\frac{V_w}{V} = \frac{\rho_b}{\rho_w}$ $\frac{V_w}{V} = \frac{800~kg/m^3}{1000~kg/m^3}$ $\frac{V_w}{V} = 0.80$ $\frac{4}{5}$ of ball 1's volume will be submerged in the water. Therefore, $\frac{1}{5}$ of ball 1 is above the surface of the water. The correct answer is: D. $\frac{1}{5}$