1. Variable Definitions

Let $S$ be the cross-sectional area of vessel B ($S=10 \text{ cm}^2$).
The area of vessel A is $\eta S$ (where $\eta=4$).
Let $y$ be the upward displacement of the piston in vessel B.
Let $h_A$ be the upward displacement of the oil level in vessel A.

2. Kinematic Constraint (Pulley System)

The string is inextensible. When the piston moves up by a distance $y$, a length of string equal to $y$ becomes slack. This slack passes over the fixed pulley to the movable pulley loop. For the movable pulley to take up a slack of length $y$, it must descend by half that distance.

Displacement of the Box ($y_{box}$):

$$ y_{box} = \frac{y}{2} $$

3. Volume Conservation

An extra mass of oil $\Delta m$ is added to vessel A. Its volume is $\Delta V = \frac{\Delta m}{\rho}$.
The total volume of liquid increases by this amount. This volume increase is distributed between the rise in level of vessel A and the rise in volume of vessel B (created by the piston moving up).

$$ \Delta V = (\text{Area}_A \cdot h_A) + (\text{Area}_B \cdot y) $$ $$ \frac{\Delta m}{\rho} = \eta S h_A + S y $$

Solving for $h_A$:

$$ h_A = \frac{1}{\eta S} \left( \frac{\Delta m}{\rho} – S y \right) \quad \dots(1) $$

4. Pressure Equilibrium

For the system to remain in equilibrium, the change in pressure at the bottom connecting tube must be equal for both arms ($\Delta P_A = \Delta P_B$).

Left Arm (A): The pressure increase is due solely to the rise in liquid level $h_A$.

$$ \Delta P_A = \rho g h_A $$

Right Arm (B): The pressure change has two components:

1. Hydrostatic pressure increase due to the column lengthening by $y$: $+ \rho g y$.
2. Change in piston pressure due to change in string tension.

Initially, the tension $2T_0 = Mg$ (weight of box).
Finally, with added mass $\Delta m$, the tension $2T’ = (M + \Delta m)g$.
The change in tension is $\Delta T = \frac{\Delta m g}{2}$.
Since tension pulls the piston up, an increase in tension decreases the pressure on the liquid below. $$ \Delta P_{piston} = -\frac{\Delta T}{S} = -\frac{\Delta m g}{2S} $$ Total pressure change at the base of tube B: $$ \Delta P_B = \rho g y – \frac{\Delta m g}{2S} $$

Equating $\Delta P_A = \Delta P_B$:

$$ \rho g h_A = \rho g y – \frac{\Delta m g}{2S} $$ $$ \rho h_A = \rho y – \frac{\Delta m}{2S} \quad \dots(2) $$

5. Solving for Displacement

Substitute Eq (1) into Eq (2):

$$ \rho \left[ \frac{1}{\eta S} \left( \frac{\Delta m}{\rho} – S y \right) \right] = \rho y – \frac{\Delta m}{2S} $$

Multiply through by $\eta S$ to clear denominators:

$$ \Delta m – \rho S y = \eta S \rho y – \frac{\eta \Delta m}{2} $$

Group terms containing $y$ on the right and $\Delta m$ on the left:

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

Solve for piston displacement $y$:

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

The displacement of the box is $y_{box} = y/2$:

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