Solution
Given Data:
- Liquid density law: $\rho = \rho_0 + ky$
- $\rho_0 = 1.0 \, \text{g/cm}^3$, $k = 0.01 \, \text{g/cm}^4$
- Ball 1: Mass $m_1 = 1.2 \, \text{g}$, Volume $V = 1.0 \, \text{cm}^3$
- Ball 2: Mass $m_2 = 1.4 \, \text{g}$, Volume $V = 1.0 \, \text{cm}^3$
- String length: $L = 15 \, \text{cm}$
Step 1: System Equilibrium
Since $m_2 > m_1$, the heavier ball (1.4 g) will be at the bottom, pulling the lighter ball (1.2 g) down. The string will be taut. We can treat the two balls and the string as a single system.
For equilibrium, the total Weight must equal the total Buoyant Force:
$$ W_{total} = F_{B1} + F_{B2} $$ $$ (m_1 + m_2)g = (V \rho_1 g) + (V \rho_2 g) $$ $$ m_1 + m_2 = V(\rho_1 + \rho_2) $$Step 2: Defining Densities
Let the depth of the upper ball ($m_1$) be $h$. Then the depth of the lower ball ($m_2$) is $h + L$.
Using the density law $\rho = \rho_0 + ky$:
- Density at upper ball: $\rho_1 = \rho_0 + kh$
- Density at lower ball: $\rho_2 = \rho_0 + k(h + L)$
Step 3: Substitution and Calculation
Substitute the density expressions into the equilibrium equation:
Now, substitute the numerical values:
$$ 1.2 + 1.4 = 1.0 \times [ 2(1.0) + 2(0.01)h + (0.01)(15) ] $$ $$ 2.6 = 2 + 0.02h + 0.15 $$ $$ 2.6 = 2.15 + 0.02h $$ $$ 0.45 = 0.02h $$ $$ h = \frac{0.45}{0.02} = 22.5 \, \text{cm} $$Step 4: Final Depths
Depth of upper ball ($m_1$) = $h = 22.5 \, \text{cm}$.
Depth of lower ball ($m_2$) = $h + L = 22.5 + 15 = 37.5 \, \text{cm}$.
Answer: The balls settle at depths of 22.5 cm and 37.5 cm.