Answer
The correct answer is:
D. $\frac{1}{5}$
Work Step by Step
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. $\frac{1}{5}$
