Problem 8 - Master Resource Book in JEE Main Physics

Question

The U-tube has a uniform cross-section as shown in figure. A liquid is filled in the two arms upto heights \(h_{1}\) and \(h_{2}\) and then the liquid is allowed to move. Neglect viscosity and surface tension. When the level equalize in the two arms, the liquid will (a) be at rest (b) be moving with an acceleration of \(g\left(\frac{h_{1}-h_{2}}{h_{1}+h_{2}+2}\right)\) (c) be moving with a velocity of \(\left(h_{1}-h_{2}\right) \sqrt{\frac{g}{2\left(h_{1}+h+h\right)}}\) (d) exert a net force to the right on the cube

Step-by-Step Solution

Answer

The liquid moves with a velocity of \((h_1-h_2)\sqrt{\frac{g}{2(h_1+h_2+2)}}\) as described in option (c).

Step 1: Understand the Problem

We have a U-tube filled with a liquid to different heights in each arm: heights \( h_1 \) and \( h_2 \). We need to find the behavior of the liquid when levels equalize, considering no viscosity or surface tension effects.

Step 2: Define Equalization of Levels

To equalize the levels in both arms, the potential energy difference due to the difference in liquid heights \((h_1 - h_2)\) initially exists. As levels equalize, this potential energy converts into kinetic energy.

Step 3: Derive Velocity from Energy Conservation

Using energy conservation, the potential energy \( mgh \) (where \( h = h_1 - h_2 \)) equals kinetic energy \( \frac{1}{2}mv^2 \). Thus, \( mgh = \frac{1}{2}mv^2 \) leads to \( v = \sqrt{2gh} \). But note, the correct usage depends on how energy distributes along the column length in U-tube.

Step 4: Analyze Given Options

We are given options for how the liquid moves. Option (c) suggests a formula for velocity considering the effective height difference and tube characteristics, with a final expression considered, \( v = (h_1 - h_2)\sqrt{\frac{g}{2(h_1 + h_2 + 2)}} \).

Step 5: Evaluate The Correctness of the Options

For this scenario, the liquid equalizes by oscillation due to its initial potential energy difference. Option (c) correctly describes the expected steady velocity of the liquid using height difference \((h_1-h_2)\). The arrangement suggests relevant distance and distribution factor \((h_1 + h_2 + 2)\) in calculation.

Key Concepts

U-tube oscillations

U-tube oscillations occur when the liquid in a U-tube moves back and forth. This happens because of the height difference between the liquid levels in each arm when the liquid is initially disturbed. Imagine the liquid in a U-shaped container. If one side has more liquid than the other, the liquid will try to balance out by moving back and forth. These oscillations are similar to what happens with a pendulum or a swinging child on a swing. The liquid moves until the gravitational forces manage to balance it out, that is, until both sides have equal heights. Time taken for these oscillations depends on how much liquid initially separates each side. A fascinating aspect of U-tube oscillations is how smoothly the liquid moves because we are neglecting friction caused by viscosity or surface tension. Therefore, the liquid’s movement is only influenced by gravity and the height differences.

Energy conservation in fluids

In fluid dynamics, energy conservation helps us understand how fluids like water move. The principle states that energy isn't lost but instead changes forms. In a meaningful sense, it focuses on potential and kinetic energies. In our U-tube scenario, initially when the liquid is not at rest, there is a height difference between the two sides of the U-tube. This difference means potential energy exists due to the gravitational pull on the higher liquid levels. As the liquid strives to balance itself, the potential energy gets converted into another form—kinetic energy, which is the energy of motion. During the oscillation, as liquids move up and down in the U-tube, some energy is swapped between these two forms until full equalization occurs. Understanding this process helps explain why fluids move and behave the way they do in dynamic environments.

Potential and kinetic energy conversion

One of the neat aspects of fluid dynamics is how potential energy and kinetic energy trade places as the liquid moves. In the U-tube example, initially, the liquid has higher potential energy at higher heights. This is like a parked roller coaster at the top of a hill with lots of stored energy waiting to change into motion. When the liquid begins to move, this potential energy starts turning into kinetic energy. Kinetic energy is the energy due to motion, and for a moving liquid, it's what allows it to flow through the U-tube. What's crucial is the mass of the liquid and the height difference, as these determine how much potential energy is initially available and consequently, how much kinetic energy can be generated. The formula for kinetic energy— \[ KE = \frac{1}{2}mv^2 \] —helps show how this energy depends on the velocity of the liquid. As the liquid velocity reaches its peak through oscillation, the energy shifts back and forth from potential to kinetic until the liquid levels equalize.