Solution to Question 10
Consider the U-tube containing three liquids A, B, and C.
Initial Equilibrium (Piston held):
- Left Arm: Liquid A (density $\rho_A$) rests on Liquid C. Total height of A is $h$.
- Right Arm: Liquid B (density $\rho_B$) rests on Liquid C.
- Connecting Tube: A horizontal tube connects the two arms. On the left, it is located at the middle of column A (depth $h/2$ from the top). On the right, it connects to liquid B.
When the piston is released, it moves to the left until equilibrium is re-established. Let $x$ be the rise in the top level of liquid A. Let $y$ be the depression of the interface between liquids A and C.
Volume Conservation:
As the piston moves, it displaces a certain volume of liquid. Assuming the arms have uniform cross-sectional area $S$:
- Left Arm: Liquid A is pushed in. The total length of the A column increases. The top rises by $x$ and the bottom pushes C down by $y$. The increase in column length is $x+y$.
- Right Arm: Liquid B is sucked out. The top level falls by $x$ and the interface with C rises by $y$ (since C is incompressible and moves from left to right). The decrease in column length is $x+y$.
At the final equilibrium, the pressure on both sides of the piston (inside the connecting tube) must be equal.
Left Side (Liquid A):
The tube is fixed at a specific physical height. Originally, the depth of the tube from the top of A was $h/2$. Since the top level of A has risen by $x$, the new depth of the tube from the surface is:
$$ \text{Depth}_A = \frac{h}{2} + x $$
$$ P_{Left} = P_{atm} + \rho_A g \left(\frac{h}{2} + x\right) $$
Right Side (Liquid B):
Originally, the height of the B column was $H_B$. The tube is at the same physical level. Since the tube was at height $h/2$ from the bottom of A, and the bottoms were level, the tube is at height $h/2$ from the bottom of B.
Original depth of tube in B = $H_B – \frac{h}{2}$.
The top level of B has fallen by $x$. Therefore, the new depth is reduced by $x$:
$$ \text{Depth}_B = \left(H_B – \frac{h}{2}\right) – x $$
$$ P_{Right} = P_{atm} + \rho_B g \left(H_B – \frac{h}{2} – x\right) $$
Equating $P_{Left} = P_{Right}$: $$ \rho_A \left(\frac{h}{2} + x\right) = \rho_B \left(H_B – \frac{h}{2} – x\right) $$ Substituting $H_B = \frac{\rho_A}{\rho_B}h$ (from initial equilibrium): $$ \rho_A \frac{h}{2} + \rho_A x = \rho_B \left( \frac{\rho_A}{\rho_B}h – \frac{h}{2} – x \right) $$ $$ \frac{\rho_A h}{2} + \rho_A x = \rho_A h – \frac{\rho_B h}{2} – \rho_B x $$ $$ (\rho_A + \rho_B)x = \rho_A h – \frac{\rho_A h}{2} – \frac{\rho_B h}{2} $$ $$ (\rho_A + \rho_B)x = \frac{h}{2}(\rho_A – \rho_B) $$ $$ \mathbf{x = \frac{h(\rho_A – \rho_B)}{2(\rho_A + \rho_B)}} $$
Now consider the pressure balance at the bottom where liquid C connects the two arms. We balance the pressure at the horizontal level of the new lower interface (the interface in the left arm, which is at depth $y$ below the original interface level).
Left Arm Pressure: $$ P_L = P_{atm} + \rho_A g (\text{Total Height of A}) $$ New total height of A = Initial $h$ + Top Rise $x$ + Bottom Dip $y$. $$ P_L = P_{atm} + \rho_A g (h + x + y) $$
Right Arm Pressure (at same level):
$$ P_R = P_{atm} + \rho_B g (\text{New Height of B}) + \rho_C g (\text{Height difference of C}) $$
New height of B = $H_B – (x + y)$.
The difference in C levels is $2y$ (left went down $y$, right went up $y$).
$$ P_R = P_{atm} + \rho_B g (H_B – x – y) + \rho_C g (2y) $$
Equating $P_L = P_R$: $$ \rho_A (h + x + y) = \rho_B (H_B – x – y) + 2\rho_C y $$ $$ \rho_A h + \rho_A(x + y) = \rho_B H_B – \rho_B(x + y) + 2\rho_C y $$ Since $\rho_A h = \rho_B H_B$ (initial equilibrium), these terms cancel: $$ \rho_A(x + y) = -\rho_B(x + y) + 2\rho_C y $$ $$ (\rho_A + \rho_B)(x + y) = 2\rho_C y $$ $$ (\rho_A + \rho_B)x + (\rho_A + \rho_B)y = 2\rho_C y $$ $$ (\rho_A + \rho_B)x = (2\rho_C – \rho_A – \rho_B)y $$ $$ y = x \cdot \frac{\rho_A + \rho_B}{2\rho_C – \rho_A – \rho_B} $$
Substitute the value of $x$ derived in Step 3 into the expression for $y$: $$ y = \left[ \frac{h(\rho_A – \rho_B)}{2(\rho_A + \rho_B)} \right] \cdot \frac{\rho_A + \rho_B}{2\rho_C – \rho_A – \rho_B} $$ The term $(\rho_A + \rho_B)$ cancels out. $$ \mathbf{y = \frac{h(\rho_A – \rho_B)}{2(2\rho_C – \rho_A – \rho_B)}} $$
Result:
Change in top level: $x = \frac{h(\rho_A – \rho_B)}{2(\rho_A + \rho_B)}$
Change in interface level: $y = \frac{h(\rho_A – \rho_B)}{2(2\rho_C – \rho_A – \rho_B)}$