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Chapter 6: Problem 30

A U-tube of base length ' \(l\) ' filled with the same volume of two liquids of densities \(\rho\) and \(2 \rho\) is moving with an acceleration ' \(a\) ' on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height \(h\) is given by (1) \(\frac{a}{2 g} l\) (2) \(\frac{3 a}{2 g} l\) (3) \(\frac{a}{g} l\) (4) \(\frac{2 a}{3 g} l\)

Short Answer

Expert verified
The height \( h \) is \( \frac{a}{g} l \), which corresponds to option (3).

Step by step solution

01

Understand the Problem

We have a U-tube with two different liquids of densities \( \rho \) and \( 2\rho \) resting in it. The tube moves horizontally with an acceleration \( a \). We need to find the height \( h \) of the liquid in one arm of the U-tube where the height difference between the surfaces becomes zero.
02

Apply the Concept of Accelerated Fluids

When a container with fluid is accelerated horizontally, the free surface of the fluid tilts. In equilibrium, the effective force per unit volume due to acceleration becomes equal to the difference in hydrostatic pressure of liquids.
03

Set Up the Pressure Balance Equation

For equilibrium, the pressure difference due to height \( h \) must equal the horizontal effective pressure difference: \[\rho g h - (2\rho g h) = \rho a l - (2\rho a l)\]
04

Simplify the Equations

Simplifying the equation, \[\rho g h - 2\rho g h = \rho a l - 2\rho a l\]This simplifies to \[ -g \rho h = -a \rho l\] which leads to \[h = \frac{a}{g} l.\]
05

Evaluate and Confirm

The calculated height \( h \) is \( \frac{a}{g} l \). This matches option (3) in the problem statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

U-tube
A U-tube is a type of container shaped like the letter "U". Its design helps to demonstrate how different fluids interact with each other and respond to various forces. In the context of our exercise, a U-tube is filled with two distinct liquids of different densities. The U-tube is often used in experiments to observe fluid behavior under different conditions, such as acceleration or changes in pressure.
  • U-tubes are widely used in practical applications due to their ability to easily show level differences and reactions to external forces.
  • They form the basis for many basic experiments in fluid dynamics.

This setup helps in visualizing concepts such as pressure differences and equilibrium under accelerated conditions.
Pressure Balance
Pressure balance is the concept of maintaining equilibrium where the forces in a system are balanced. When working with fluids, pressure balance helps determine how pressures exerted by fluids in different sections should equalize.
  • In the U-tube, the different densities and acceleration require careful balance to achieve equilibrium.
  • For our scenario, pressure balance involves horizontal forces due to acceleration equalizing the differences in hydrostatic pressure across the heights of the fluids.

The pressure balance equation we used in the exercise is crucial. It involves equating the effective pressure due to acceleration with the pressure difference created by liquid heights.
Fluid Dynamics
Fluid dynamics studies how fluids behave and flow, especially when subjected to forces. This is crucial for understanding accelerated fluids in our U-tube scenario. When the tube moves horizontally with acceleration, the fluids adjust their surfaces to establish equilibrium.
  • Fluid dynamics involves predicting the tilting and levels of fluids based on applied forces.
  • Understanding fluid dynamics helps in setting up equations to resolve changes in fluid surfaces under acceleration.

This helps explain how the freely moving interfaces between different fluids can end up level when the entire system is equally balanced against external forces.
Hydrostatic Pressure
Hydrostatic pressure refers to the pressure exerted by a fluid at rest due to the force of gravity. It's calculated based on the height of the fluid column and the fluid density. In the U-tube exercise, understanding hydrostatic pressure is essential for setting up the balance equations.
  • Hydrostatic pressure depends directly on both the fluid's density and its height.
  • It plays a key role in determining how much pressure is exerted on a fluid's surface at any depth.

By knowing the hydrostatic pressure, you can deduce the difference in levels between two sections of the U-tube, leading to equilibrium and negating any height differences in the accelerated tube.

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Most popular questions from this chapter

Bernoulli's equation can be written in the following different forms (Column I). Column II lists certain units each of which pertains to one of the possible forms of the equatioa. Match the unit associated with each of the equations. $$ \begin{array}{|l|l|} \hline \text { Column I } & {2}{|c|} {\text { Column II }} \\ \hline \text { i. } \frac{v^{2}}{2 g}+\frac{p}{\rho g}+z=\text { constant } & \begin{array}{l} \text { a. Total energy } \\ \text { per unit mass } \end{array} \\ \hline \text { ii. } \frac{\rho v^{2}}{2}+P+\rho g z=\text { constant } & \begin{array}{l} \text { b. Total energy } \\ \text { per unit weight } \end{array} \\ \hline \text { iii. } \frac{v^{2}}{2}+\frac{P}{\rho}+g z=\text { constant } & \begin{array}{r} \text { c. Total energy } \\ \text { per unit volume } \end{array} \\ \hline \end{array} $$

A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross section \(A_{1}\) and \(A_{2}\), are \(v_{1}\) and \(v_{2}\), respectively. The difference in the levels of the liquid in the two vertical tubes is \(h\). Then (1) the volume of the liquid flowing through the tube in unit time is \(A_{1} v_{1}\) (2) \(v_{2}-v_{1}=\sqrt{2 g h}\) (3) \(v_{2}^{2}-v_{1}^{2}=2 g h\) (4) the energy per unit mass of the liquid is the same in both sections of the tube

For Problem 29-30 In a U-tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. In a U-tube \(20 \mathrm{~cm}\) of a liquid of density \(\rho\) is on left hand side and \(10 \mathrm{~cm}\) of another liquid of density \(1.5 \rho\) is on right hand side. In between them there is a third liquid of density \(2 \rho\). What is the value of \(h ?\) (1) \(5 \mathrm{~cm}\) (2) \(2.5 \mathrm{~cm}\) (3) \(2 \mathrm{~cm}\) (4) \(7.5 \mathrm{~cm}\)

A ball of density \(\rho\) is dropped onto a horizontal solid surface. It bounces elastically from the surface and returns to its original position in time \(t_{1}\). Next, the ball is released from the same height, but this time it strikes the surface of a liquid of density \(\rho_{L}(>\rho\) but less than \(2 \rho)\), and takes \(t\) : second to come back to its original height (1) \(t_{2}>t_{1}\) (2) \(t_{1}>t_{2}\) (3) the motion of the ball is not simple harmonic (4) If \(\rho=\rho_{L}\), then the speed of the ball inside the liquid will be independent of its depth

A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side until water rises by \(25 \mathrm{~cm}\) on the other side. If the density of oil is \(0.8\), the oil level will stand higher than the water level by (1) \(6.25 \mathrm{~cm}\) (2) \(12.50 \mathrm{~cm}\) (3) \(31.75 \mathrm{~cm}\) (4) \(62.50 \mathrm{~cm}\)
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