tHERMAL O15

Solution for Question 15

Initial Analysis:

Initially, both pistons are at the same height (40 cm) and the system is in equilibrium. The pressure must be equal in both cylinders.

$$P = P_0 + \frac{m_A g}{A_A} = P_0 + \frac{m_B g}{A_B}$$

Given $m_A = 1 \text{ kg}$ and $m_B = 2 \text{ kg}$.

$$\frac{1}{A_A} = \frac{2}{A_B} \Rightarrow A_B = 2A_A$$

The Disturbance:

An additional mass of 10g is gently placed on A.

The pressure requirement to support Piston A becomes higher than Piston B. Piston A will move down, pushing gas into Cylinder B, making Piston B move up.

Final State Dynamics:

Since the pressure required to support A (now heavier) is permanently higher than that required to support B, equilibrium cannot be restored in the middle. Piston A will sink all the way to the bottom.

Volume Calculation:

Piston A moves down by $h_A = 40 \text{ cm}$.

Volume pushed out = $A_A \times 40$.

This volume moves to B. Piston B moves up by $h_B$.

$$A_B \times h_B = A_A \times 40$$

Since $A_B = 2A_A$:

$$2A_A \times h_B = 40 A_A \Rightarrow h_B = 20 \text{ cm}$$

Results:

1. Position: A is at 0 cm. B is at $40 + 20 = 60 \text{ cm}$. Height difference = 60 cm.
2. Pressure: Once A hits the bottom, the gas only needs to support Piston B (which has not changed mass). Therefore, the final gas pressure returns to the initial value required to support B.