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Chapter 6: Problem 45

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) \(62.50 \mathrm{~cm}\)

Short Answer

Expert verified
The oil level will stand 6.25 cm higher than the water level.

Step by step solution

01

Understand the concept

This problem involves a U-tube partially filled with different liquids (water and oil). The densities of these liquids will determine the balance of their levels. We will use the principle that the pressure at the same height in a fluid is equal, i.e., the pressure of the oil column at a certain height will equal the pressure of the water column at the same height.
02

Apply hydrostatic pressure equation to balance points

Hydrostatic pressure is given by the formula \( P = h \cdot \rho \cdot g \), where \( h \) is the height of the fluid column, \( \rho \) is the density, and \( g \) is the acceleration due to gravity. Since water rises by 25 cm on one side, we need to find the height of the oil column that balances this water column. The formula for pressure can be set equal for both sides:\[ h_o \cdot \rho_o \cdot g = 25 \cdot \rho_w \cdot g \]where \( h_o \) is the height of the oil, \( \rho_o = 0.8 \) (density of oil), and \( \rho_w = 1 \) (density of water).
03

Solve the equation for oil height

Cancel the gravitational acceleration \( g \) from both sides since it appears in both terms. Rearranging the equation gives us: \[ h_o = \frac{25 \times 1}{0.8} \]This simplifies to \( h_o = 31.25 \text{ cm} \).
04

Calculate how much higher the oil level stands

Since the water rose by 25 cm, we subtract this height from the oil height to find out how much higher the oil column stands above the original water level:\[ 31.25 - 25 = 6.25 \text{ cm} \].Thus, the oil level will be 6.25 cm higher than the water level.

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Key Concepts

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

Density: The Mass Within Volume
Density is a measure of how much mass is contained within a certain volume. It's an important factor in determining how different fluids interact with each other.
  • Density is represented by the symbol \( \rho \).
  • It is calculated using the formula: \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume.
  • High-density substances are heavier for the same volume compared to low-density substances.
In the context of a U-tube scenario, understanding density helps predict how liquids like oil and water will balance out. The density of oil (0.8) is less than water (1), which causes it to rise higher for the same pressure in a U-tube setup.
Grasping density differences is essential for explaining why one liquid can "float" above another, leading to a varied height of liquid columns in our exercise.
Fluid Column: Balancing Act
A fluid column refers to a vertical section of fluid within a container, like water or oil sitting in a U-tube. This section of liquid holds weight due to gravity and exerts pressure downward.
  • The height of a fluid column is directly proportional to the pressure it exerts.
  • In a connected system like a U-tube, different fluids achieve balance by matching their pressures at corresponding heights.
This balance is achieved regardless of their individual volumes or density contrasts. The pressure exerted by each column results from its density and height.
Understanding fluid columns is pivotal in this exercise as it helps explain why the oil needed to rise higher than water to equate their pressures.
U-tube: The Vessel of Equilibrium
A U-tube is a simple device that holds liquids in its curved tubes, resembling the shape of the letter "U". It is commonly used to demonstrate principles of fluid mechanics.
  • The key characteristic of a U-tube is its ability to show the equilibrium between two liquid columns.
  • Unequal densities will still balance at different heights, providing a visual demonstration of hydrostatic principles.
In our exercise, when oil is added to one side, the water on the other rises. This shift of liquid levels inside the U-tube showcases how different densities alter the heights of fluid columns.
An understanding of U-tube mechanics can enhance comprehension of balancing pressures, which is critical to solving hydrostatic problems.
Acceleration Due to Gravity: The Force on Fluids
Acceleration due to gravity, symbolized as \( g \), plays a crucial role in hydrostatics as it affects the pressure exerted by a fluid column.
  • On Earth, gravity pulls at an approximate rate of \( 9.8 \text{ m/s}^2 \).
  • This force causes the fluid to exert pressure at its base, calculated by \( P = h \cdot \rho \cdot g \).
In our U-tube example, gravity is the driving force that brings water and oil to equilibrium despite their density differences.
Cancellation of \( g \) in pressure equations emphasizes that regardless of gravity鈥檚 force, it鈥檚 the balance of pressure between fluids with different densities that determines equilibrium in this scenario. Understanding gravity鈥檚 role helps clarify how pressure dynamics play out in practical applications.

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Most popular questions from this chapter

The area of two holes \(A\) and \(B\) are \(2 a\) and \(a\), respectively. The holes are at height \((H / 3)\) and \((2 H / 3)\) from the surface of water. Find the correct option(s): (1) The velocity of efflux at hole \(B\) is 2 times the velocity of efflux at hole \(A\). (2) The velocity of efflux at hole \(B\) is \(\sqrt{2}\) time the velocity of efflux at hole \(A\). (3) The discharge is same through both the holes. (4) The discharge through hole \(A\) is \(\sqrt{2}\) time the discharge through hole \(B\)

Water is floating smoothly through a closed-pipe system. At one point \(A\), the speed of the water is \(3.0 \mathrm{~m} / \mathrm{s}\) while at another point \(B, 1.0 \mathrm{~m}\) higher, the speed is \(4.0 \mathrm{~m} / \mathrm{s}\). The pressure at \(A\) is \(20 \mathrm{kPa}\) when the water is flowing and \(18 \mathrm{kPa}\) when the water flow stops. Then (1) the pressure at \(B\) when water is flowing is \(6.5 \mathrm{kPa}\) (2) the pressure at \(B\) when water is flowing is \(8.0 \mathrm{kPa}\) (3) the pressure at \(B\) when water stops flowing is \(10 \mathrm{kPa}\) (4) the pressure at \(B\) when water stops flowing is \(8.0 \mathrm{kPa}\)

A cylindrical block is floating (partially submerged) in a vessel containing water. Initially, the platform on which the vessel is mounted is at rest. Now the platform along with the vessel is allowed to fall freely under gravity. As a result, the buoyancy force (1) becomes zero (2) decreases (3) increases (4) information is insufficient

Length of horizontal arm of a uniform cross section \(U\)-tube is \(l=21 \mathrm{~cm}\) and ends of both of the vertical arms are open to surrounding of pressure \(10 \times 500 \mathrm{~N} / \mathrm{m}^{2}\). A liquid of density \(\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) is poured into the tube such that liquid just fillg is sealed and the tube is then rotated about a vertical axiy arm with angular val \(\omega_{0}=10 \mathrm{rad} / \mathrm{s} .\) If length of each vertical arm is \(a=6 \mathrm{~cm}\) calculate the length of air column in the sealed arm. (in cm).

A horizontal tube has different cross sections at points \(A\) and B. The areas of cross section are \(a_{1}\) and \(a_{2}\), respectively, and pressures at these points are \(p_{1}=\rho g h_{1}\) and \(p_{2}=\rho g h_{2}\), where \(\rho\) is the density of liquid flowing in the tube and \(h_{1}\) and \(h_{2}\) are heights of liquid columns in vertical tubes connected at \(A\) and \(B\). If \(h_{1}-h_{2}=h\), then the flow rate of the liquid in the horizontal tube is (1) \(a_{1} a_{2} \sqrt{\frac{2 g h}{a_{1}^{2}-a_{2}^{2}}}\) (2) \(a_{1} a_{2} \sqrt{\frac{2 g}{h\left(a_{1}^{2}-a_{2}^{2}\right)}}\) (3) \(a_{1} a_{2} \sqrt{\frac{\left(a_{1}^{2}+a_{2}^{2}\right) h}{2 g\left(a_{1}^{2}-a_{2}^{2}\right)}}\) (4) \(\frac{2 a_{1} a_{2} g h}{\sqrt{a_{1}^{2}-a_{2}^{2}}}\)
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