open glass capil ered into a cis nsity = 13600 kg
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
![**Capillary Depression of Mercury in a Glass Tube**
_When an open glass capillary tube with a bore of 2 mm is lowered into a cistern containing mercury, various factors influence the behavior of the mercury within the tube. The physical properties of mercury and the interaction between the mercury and glass play a critical role in determining the depression level._
### Parameters Provided:
- **Density of Mercury (ρ)**: 13,600 kg/m³
- **Contact Angle between Mercury and Glass (θ)**: 140°
- **Surface Tension Coefficient (γ)**: 0.484 N/m
- **Gravitational Acceleration (g)**: 9.81 m/s²
- **Bore Diameter of the Capillary Tube (d)**: 2 mm (which translates to a radius \( r \) of 1 mm or 0.001 m)
### Objective:
Calculate the depression of mercury in the capillary tube below the free surface in the cistern. The depression (h) can be obtained using the formula:
\[ h = \frac{2 \gamma \cos \theta}{\rho g r} \]
Where:
- \( γ \) is the surface tension coefficient.
- \( θ \) is the contact angle between the liquid and the surface.
- \( ρ \) is the density of the liquid.
- \( g \) is the acceleration due to gravity.
- \( r \) is the radius of the tube.
**Note**: Since mercury depresses in the glass tube (as opposed to water rising in a capillary tube), the expected result is a negative value indicating depression.
### Calculation:
1. Convert the bore diameter to radius:
\[ r = \frac{d}{2} = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 0.001 \text{ m} \]
2. Substitute the known values into the formula:
\[ h = \frac{2 \cdot 0.484 \text{ N/m} \cdot \cos(140°)}{13600 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2 \cdot 0.001 \text{ m}} \]
### Visual Aids:
If needed, diagrams or graphs illustrating:
1. **Capillary Depression**:
- A glass tube partially](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7726c74f-91d8-429b-86a6-10eef1db1015%2F56d5780a-cc5d-48fa-a7b2-eaeb3b1a223f%2F87ysou9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Capillary Depression of Mercury in a Glass Tube**
_When an open glass capillary tube with a bore of 2 mm is lowered into a cistern containing mercury, various factors influence the behavior of the mercury within the tube. The physical properties of mercury and the interaction between the mercury and glass play a critical role in determining the depression level._
### Parameters Provided:
- **Density of Mercury (ρ)**: 13,600 kg/m³
- **Contact Angle between Mercury and Glass (θ)**: 140°
- **Surface Tension Coefficient (γ)**: 0.484 N/m
- **Gravitational Acceleration (g)**: 9.81 m/s²
- **Bore Diameter of the Capillary Tube (d)**: 2 mm (which translates to a radius \( r \) of 1 mm or 0.001 m)
### Objective:
Calculate the depression of mercury in the capillary tube below the free surface in the cistern. The depression (h) can be obtained using the formula:
\[ h = \frac{2 \gamma \cos \theta}{\rho g r} \]
Where:
- \( γ \) is the surface tension coefficient.
- \( θ \) is the contact angle between the liquid and the surface.
- \( ρ \) is the density of the liquid.
- \( g \) is the acceleration due to gravity.
- \( r \) is the radius of the tube.
**Note**: Since mercury depresses in the glass tube (as opposed to water rising in a capillary tube), the expected result is a negative value indicating depression.
### Calculation:
1. Convert the bore diameter to radius:
\[ r = \frac{d}{2} = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 0.001 \text{ m} \]
2. Substitute the known values into the formula:
\[ h = \frac{2 \cdot 0.484 \text{ N/m} \cdot \cos(140°)}{13600 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2 \cdot 0.001 \text{ m}} \]
### Visual Aids:
If needed, diagrams or graphs illustrating:
1. **Capillary Depression**:
- A glass tube partially
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