Problem 74 - Fundamentals of Physics

Question

A simple open U-tube contains mercury. When \(11.2 \mathrm{~cm}\) of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

Step-by-Step Solution

Answer

The mercury rises approximately 0.82 cm in the left arm.

Step 1: Understanding the Problem

We have a U-tube partially filled with mercury, and we add water to one arm. We need to find out how high the mercury rises in the other arm. This involves understanding the principle of balancing pressures in fluids.

Step 2: Identify Given Values

Let's identify the key values needed for this problem. The height of water added is given as \(11.2 \text{ cm}\). We also know that the density of mercury is much higher than that of water.

Step 3: Apply Pressure Balance

In a U-tube, the weight of the water column in one arm creates pressure that causes the mercury to rise in the other arm. The pressure exerted by the water is equal to the additional height of mercury displaced, but remember that mercury's density affects how much it rises.

Step 4: Use Pressure Formula

We can use the formula for pressure balance. Pressure due to water column = Pressure due to mercury column: \( \rho_{\text{water}} \cdot g \cdot h_{\text{water}} = \rho_{\text{mercury}} \cdot g \cdot h_{\text{mercury}} \).The \( g \) cancels out and the densities \( \rho \) are used to find the rise in mercury.

Step 5: Substitute Known Values

We substitute the known values into the equation. Let \( \rho_{\text{water}} = 1 \text{ g/cm}^3\) and \( \rho_{\text{mercury}} = 13.6 \text{ g/cm}^3\). We find that:\[ 1 \cdot 11.2 = 13.6 \cdot h_{\text{mercury}} \] Solving for \( h_{\text{mercury}} \), we find it is \( \frac{11.2}{13.6} \text{ cm} \).

Step 6: Calculate the Result

Now, calculate the rise of mercury:\[ h_{\text{mercury}} = \frac{11.2}{13.6} \approx 0.8235 \text{ cm} \].This means that the mercury rises about \( 0.82 \text{ cm} \) on the left side.

Key Concepts

Pressure Balance

When working with fluids in containers like a U-tube, understanding the concept of pressure balance is essential. Pressure in a fluid is transmitted uniformly in all directions. In the case of the U-tube, when water is added to one side, it creates a pressure difference. This difference makes the mercury rise in the other side.
\[ \rho_{ ext{water}} \cdot g \cdot h_{ ext{water}} = \rho_{ ext{mercury}} \cdot g \cdot h_{ ext{mercury}} \]
This equation helps to understand how the height of water affects the height of mercury. Despite the different densities, pressures must be balanced in both arms of the U-tube. This means that the product of density, gravity, and height is constant across a connected fluid system like this.

Density

Density, denoted usually as \( \rho \), plays a crucial role in fluid dynamics. It is a measure of mass per unit volume and is expressed in \( \text{g/cm}^3 \) for this problem. Mercury, with a density of \(13.6 \text{ g/cm}^3\), is much denser than water, which has a density of \(1 \text{ g/cm}^3\).
This large difference in density dictates how the fluids behave in the U-tube. A dense fluid like mercury needs a smaller height to exert the same pressure as a much taller column of water.

• Density impacts the fluid's buoyancy and pressure exerted on container walls.
• When calculating pressure differences, remember to account for the density difference between the interacting fluids.

U-tube

A U-tube is a helpful device to visualize fluid interactions and pressure concepts. It is shaped like the letter "U" and can hold different fluids in each arm. When a fluid is added to one side, it exerts pressure due to its weight, causing the fluid level in the other side to change.

• The U-tube setup allows for straightforward experimentation and calculation of fluid behaviors.
• In this problem, the U-tube helps demonstrate how mercury rises when water is added to the opposite side due to pressure balance.
• The U-shape ensures that both arms are at the same atmospheric pressure, simplifying pressure calculations based on the liquid columns alone.

Mercury-Water Interaction

The interaction between mercury and water in a U-tube is fascinating because of their contrasting densities. Mercury, being very dense, doesn't rise as much in response to the water being added on the other side.
This specific interaction highlights several points:

• Even a small addition of water with much less density can move denser mercury significantly, due to pressure balance.
• The phenomenon shows how heavy, dense fluids like mercury respond differently to pressure than lighter fluids like water.
• Understanding mercury-water interaction provides insights into predicting fluid behaviors in various engineering and scientific applications.

Such insights are important when applying fluid dynamics concepts to real-world scenarios.