1. Kinematic Analysis (Constraint Relations)

Let the box move down by a distance $y$. The box is attached to the movable pulley.

For the movable pulley to descend by $y$, the length of the string segment supporting it must increase. Since one end of the string is fixed to the ceiling, the slack must come from the piston side.

From standard pulley constraints, if the movable pulley moves down by $y$, it pulls the rope from the free end by a distance $2y$. Therefore, Piston B moves UP by $2y$.

2. Fluid Volume Conservation

As Piston B moves up, fluid flows from Vessel A to Vessel B (or levels adjust). Let Piston B move up by $y_B = 2y$.

Volume swept by Piston B = $S_B \cdot y_B = S_B(2y)$.

This volume comes from Vessel A. Since Vessel A has a cross-sectional area $S_A = \eta S_B$, the drop in level of A ($y_A$) is:

$$ y_A = \frac{S_B \cdot 2y}{S_A} = \frac{S_B \cdot 2y}{\eta S_B} = \frac{2y}{\eta} $$

So, the oil level in A drops by $2y/\eta$.

3. Addition of Mass

We add mass $\Delta m$ (sand) to the box and mass $\Delta m$ (oil) to Vessel A.

• Effect of Oil: Adding oil of mass $\Delta m$ to vessel A raises the level of A by: $$ h_{oil} = \frac{\Delta m}{\rho S_A} $$
• Effect of Sand: Adding sand to the box increases the tension in the string. $$ 2\Delta T = \Delta m g \implies \Delta T = \frac{\Delta m g}{2} $$ This tension increase acts as an upward force on Piston B, effectively reducing the pressure required below it (or pulling it up). The pressure change equivalent at piston B is: $$ \Delta P_{tension} = \frac{\Delta T}{S_B} = \frac{\Delta m g}{2 S_B} $$

4. Pressure Balance Equation

For equilibrium at the connecting tube level, the pressure from the A-side must equal the pressure from the B-side.

Change in Pressure at A: The level dropped by $2y/\eta$ due to piston movement and rose by $h_{oil}$ due to added oil.

$$ \Delta P_A = \rho g \left( h_{oil} – \frac{2y}{\eta} \right) = \rho g \left( \frac{\Delta m}{\rho S_A} – \frac{2y}{\eta} \right) $$

Change in Pressure at B: The piston B moved up by $2y$. Also, the tension pulled it up (reducing pressure). The fluid pressure must increase to support the new height $2y$ minus the help from tension.

$$ \Delta P_B = \rho g (2y) – \frac{\Delta m g}{2 S_B} $$

Equating $\Delta P_A = \Delta P_B$:

$$ \rho g \left( \frac{\Delta m}{\rho S_A} – \frac{2y}{\eta} \right) = \rho g (2y) – \frac{\Delta m g}{2 S_B} $$

Divide by $g$ and rearrange to solve for $y$. Note $S_A = \eta S_B$ and $S = S_B$.

$$ \frac{\Delta m}{S_A} – \frac{2\rho y}{\eta} = 2\rho y – \frac{\Delta m}{2 S_B} $$ $$ \frac{\Delta m}{\eta S} + \frac{\Delta m}{2S} = 2\rho y + \frac{2\rho y}{\eta} $$ $$ \frac{\Delta m}{S} \left( \frac{1}{\eta} + \frac{1}{2} \right) = 2\rho y \left( 1 + \frac{1}{\eta} \right) $$ $$ \frac{\Delta m}{S} \left( \frac{2+\eta}{2\eta} \right) = 2\rho y \left( \frac{\eta+1}{\eta} \right) $$

Solving for $y$:

$$ y = \frac{\Delta m (\eta + 2)}{4 \rho S (\eta + 1)} $$

5. Calculation

Substituting the values:

• $\Delta m = 100 \text{ g} = 0.1 \text{ kg}$
• $\eta = 4$
• $\rho = 0.5 \text{ g/cm}^3 = 500 \text{ kg/m}^3$
• $S = 10 \text{ cm}^2 = 10^{-3} \text{ m}^2$

$$ y = \frac{0.1 (4 + 2)}{4 (500) (10^{-3}) (4 + 1)} $$ $$ y = \frac{0.1 \cdot 6}{2 \cdot 5} = \frac{0.6}{10} = 0.06 \text{ m} $$ $$ y = 6.0 \text{ cm} $$