Question

The U-tube shown has a uniform cross-section. A liquid is filled in the two arms up to heights $h_1$ and $h_2$, and then the liquid is allowed to move. Neglect viscosity and surface tension. When the levels equalize in the two arms, the liquid will

• A. be at rest
• B. be moving with an acceleration of $g \left(\frac{h_1 - h_2}{h_1 + h_2 + h}\right)$
• C. be moving with a velocity of $(h_1 - h_2) \sqrt{\frac{g}{2(h_1 + h_2 + h)}}$
• D. exert a net force to the right on the tube

Answer 1

Answer: (C)

be moving with a velocity of $(h_1 - h_2) \sqrt{\frac{g}{2(h_1 + h_2 + h)}}$

Answer 2

The problem describes a liquid in a U-tube with different initial heights ($h_1$ and $h_2$) in its two arms. We need to find the state of the liquid when the levels equalize, assuming no viscosity or surface tension.

1. Identify the system and initial/final states:

• The system is the liquid column in the U-tube.
• Initial state: Liquid levels are $h_1$ and $h_2$. The liquid is at rest ($KE_i = 0$).
• Final state: Liquid levels are equal. Let this height be $h_f$. Since the total volume of the liquid is constant and the tube has a uniform cross-section, the total length of the liquid column ($h_1 + h_2 + h$) remains constant.
  When levels equalize, the liquid column will have a length $h_f$ in each vertical arm and length $h$ in the horizontal arm.
  So, $h_f + h_f + h = h_1 + h_2 + h$.
  This implies $2h_f = h_1 + h_2$, so $h_f = \frac{h_1 + h_2}{2}$.
  At this point, the liquid will be moving with some velocity $v$ (since it's an equilibrium position of an oscillation, where kinetic energy is maximum).

2. Apply the principle of conservation of mechanical energy:

Since viscosity (damping) and surface tension are neglected, the total mechanical energy (potential energy + kinetic energy) of the liquid column is conserved.
$KE_i + PE_i = KE_f + PE_f$

Let $A$ be the uniform cross-sectional area of the tube and $\rho$ be the density of the liquid.
Let's choose the bottom of the horizontal section of the U-tube as the reference level for potential energy ($PE=0$).

• Initial Potential Energy ($PE_i$):
  The mass of liquid in the left arm is $m_1 = \rho A h_1$, with its center of mass (CM) at height $h_1/2$.
  The mass of liquid in the right arm is $m_2 = \rho A h_2$, with its CM at height $h_2/2$.
  The mass of liquid in the horizontal section is $m_h = \rho A h$, with its CM at height $0$.
  $PE_i = m_1 g \frac{h_1}{2} + m_2 g \frac{h_2}{2} + m_h g \cdot 0$
  $PE_i = (\rho A h_1) g \frac{h_1}{2} + (\rho A h_2) g \frac{h_2}{2}$
  $PE_i = \frac{1}{2} \rho A g (h_1^2 + h_2^2)$

• Initial Kinetic Energy ($KE_i$):
  The liquid is initially at rest.
  $KE_i = 0$

• Final Potential Energy ($PE_f$):
  When levels equalize, the height in each arm is $h_f = \frac{h_1 + h_2}{2}$.
  The mass of liquid in the left arm is $\rho A h_f$, with CM at $h_f/2$.
  The mass of liquid in the right arm is $\rho A h_f$, with CM at $h_f/2$.
  $PE_f = (\rho A h_f) g \frac{h_f}{2} + (\rho A h_f) g \frac{h_f}{2}$
  $PE_f = \rho A g h_f^2 = \rho A g \left(\frac{h_1 + h_2}{2}\right)^2$

• Final Kinetic Energy ($KE_f$):
  When the levels equalize, the entire liquid column of total mass $M = \rho A (h_1 + h_2 + h)$ moves with a velocity $v$.
  $KE_f = \frac{1}{2} M v^2 = \frac{1}{2} \rho A (h_1 + h_2 + h) v^2$

3. Substitute into the energy conservation equation:

$0 + \frac{1}{2} \rho A g (h_1^2 + h_2^2) = \frac{1}{2} \rho A (h_1 + h_2 + h) v^2 + \rho A g \left(\frac{h_1 + h_2}{2}\right)^2$

Divide the entire equation by $\frac{1}{2} \rho A$:
$g (h_1^2 + h_2^2) = (h_1 + h_2 + h) v^2 + 2g \left(\frac{h_1 + h_2}{2}\right)^2$
$g (h_1^2 + h_2^2) = (h_1 + h_2 + h) v^2 + \frac{g}{2} (h_1 + h_2)^2$

Rearrange to solve for $v^2$:
$(h_1 + h_2 + h) v^2 = g (h_1^2 + h_2^2) - \frac{g}{2} (h_1 + h_2)^2$
$(h_1 + h_2 + h) v^2 = g \left(h_1^2 + h_2^2 - \frac{1}{2}h_1^2 - h_1h_2 - \frac{1}{2}h_2^2\right)$
$(h_1 + h_2 + h) v^2 = g \left(\frac{1}{2}h_1^2 - h_1h_2 + \frac{1}{2}h_2^2\right)$
$(h_1 + h_2 + h) v^2 = \frac{g}{2} (h_1^2 - 2h_1h_2 + h_2^2)$
$(h_1 + h_2 + h) v^2 = \frac{g}{2} (h_1 - h_2)^2$
$v^2 = \frac{g (h_1 - h_2)^2}{2(h_1 + h_2 + h)}$
$v = |h_1 - h_2| \sqrt{\frac{g}{2(h_1 + h_2 + h)}}$

Since the options provide $(h_1 - h_2)$ without an absolute value, we assume $h_1 > h_2$ or the magnitude is implied.

4. Evaluate the options:

(a) be at rest: Incorrect. At the equilibrium position (levels equalized), the potential energy is minimum, so kinetic energy must be maximum, meaning the liquid is moving. (b) be moving with an acceleration of $g \left(\frac{h_1 - h_2}{h_1 + h_2 + h}\right)$: Incorrect. The motion of the liquid in the U-tube is simple harmonic motion (SHM). In SHM, the acceleration is zero at the equilibrium position (when levels equalize). The given expression for acceleration is the initial acceleration (or maximum acceleration if $h_1-h_2$ is the amplitude). (c) be moving with a velocity of $(h_1 - h_2) \sqrt{\frac{g}{2(h_1 + h_2 + h)}}$: This matches our derived velocity. This is the maximum velocity of the liquid during its oscillation. (d) exert a net force to the right on the tube: Incorrect. Since the acceleration of the liquid column is zero when the levels equalize (it's the equilibrium point of SHM), the net force acting on the liquid column is zero. By Newton's third law, the net force exerted by the liquid on the tube is also zero.