Let $m_0$, $V_0$, and $\rho_0$ be the mass, volume, and density of the ball, respectively, such that $m_0 = \rho_0 V_0$.

[Q1] Compare $m_A$ and $m_B$ The mass of the overflown liquid is equal to the mass of the displaced liquid. In liquid A, the ball floats. According to Archimedes' principle for a floating object, the buoyant force equals the ball's weight. The weight of the displaced liquid equals the weight of the ball.

$$W_{\text{disp, A}} = W_0$$ $$m_A g = m_0 g \implies m_A = m_0$$

In liquid B, the ball sinks. This implies that the ball's density is greater than the liquid's density, i.e., $\rho_0 > \rho_B$. The volume of the displaced liquid is the entire volume of the ball, $V_0$.

$$m_B = \rho_B V_{\text{disp, B}} = \rho_B V_0$$

Comparing $m_A$ with $m_B$:

$$m_A = m_0 = \rho_0 V_0$$

Since $\rho_0 > \rho_B$, it follows that $\rho_0 V_0 > \rho_B V_0$, which means $m_A > m_B$.

[Q2] What is $\rho_B$? From the analysis in Q1, the mass of the liquid B displaced is $m_B$, and the volume displaced is $V_0$. The density of liquid B is therefore:

$$\rho_B = \frac{m_B}{V_0}$$

We can express the ball's volume $V_0$ using its mass $m_0$ and density $\rho_0$ as $V_0 = m_0/\rho_0$. From Q1, we established that $m_0 = m_A$. Substituting this in:

$$\rho_B = \frac{m_B}{m_A/\rho_0} = \rho_0 \frac{m_B}{m_A}$$

[Q3] Suppose $\frac{2}{3}$ of the ball is above the liquid A, what is $\rho_A$? For the floating ball, the buoyant force equals the ball's weight:

$$F_{B,A} = W_0$$

The buoyant force is the weight of the displaced fluid, $F_{B,A} = \rho_A V_{sub} g$, where $V_{sub}$ is the submerged volume. The ball's weight is $W_0 = \rho_0 V_0 g$. If $2/3$ of the ball is above the liquid, the submerged fraction is $1 - 2/3 = 1/3$. Thus, $V_{sub} = \frac{1}{3} V_0$. Equating the forces:

$$\rho_A \left(\frac{1}{3} V_0\right) g = \rho_0 V_0 g$$

Canceling $V_0 g$ from both sides gives:

$$\frac{1}{3}\rho_A = \rho_0 \implies \rho_A = 3\rho_0$$

[Q4] Suppose two liquid can perfectly mix together without separation, what is the minimum percentage of liquid A in the mixed liquid for the ball to float? The ball will float if its density is less than or equal to the density of the mixed liquid, $\rho_{mix} \ge \rho_0$. The minimum condition for floating is $\rho_{mix} = \rho_0$. Let $p$ be the volume fraction of liquid A in the mixture. Assuming the volumes are additive, the density of the mixture is:

$$\rho_{mix} = p \rho_A + (1-p) \rho_B$$

Setting $\rho_{mix} = \rho_0$ for the minimum condition:

$$p \rho_A + (1-p) \rho_B = \rho_0$$

Solving for $p$:

$$p(\rho_A - \rho_B) = \rho_0 - \rho_B \implies p = \frac{\rho_0 - \rho_B}{\rho_A - \rho_B}$$

Substitute the expressions for $\rho_A$ and $\rho_B$ from Q3 and Q2:

$$p = \frac{\rho_0 - \rho_0 \frac{m_B}{m_A}}{3\rho_0 - \rho_0 \frac{m_B}{m_A}} = \frac{\rho_0 \left(1 - \frac{m_B}{m_A}\right)}{\rho_0 \left(3 - \frac{m_B}{m_A}\right)} = \frac{\frac{m_A-m_B}{m_A}}{\frac{3m_A-m_B}{m_A}}$$ $$p = \frac{m_A - m_B}{3m_A - m_B}$$

The minimum percentage is $100 \times p$.