A cylinder of length `l` and radius `r` is heated to temperature `T`. A longitudeinal compressive force `F` is applied on cylinder to keep its length same. Find coefficient of volume expansion.

Options

A. `9E//(pi r^(2) YT)`
B. `6F//(pi r^(2) YT)`
C. `3F//(pi r^(2) YT)`
D. `F//(3 pi r^(2) YT)`

Answer

To find the coefficient of volume expansion of a cylinder subjected to a longitudinal compressive force while being heated, we can follow these steps:

### Step 1: Understand the Problem
We have a cylinder of length \( l \) and radius \( r \) that is heated to a temperature \( T \). A longitudinal compressive force \( F \) is applied to keep its length constant. We need to find the coefficient of volume expansion, denoted as \( \gamma \).

### Step 2: Define Stress and Strain
When the compressive force \( F \) is applied, it creates a stress in the cylinder. The stress (\( \sigma \)) is defined as:
\[ \sigma = \frac{F}{A} \]
where \( A \) is the cross-sectional area of the cylinder, given by \( A = \pi r^2 \). Thus, the stress can be expressed as:
\[ \sigma = \frac{F}{\pi r^2} \]

### Step 3: Relate Stress to Strain
The strain (\( \epsilon \)) in the material is related to the stress by Young's modulus (\( Y \)):
\[ \sigma = Y \cdot \epsilon \]
Since we are keeping the length constant, the strain can also be expressed in terms of the coefficient of linear expansion (\( \alpha \)) and the change in temperature (\( \Delta T \)):
\[ \epsilon = \alpha \Delta T \]

### Step 4: Equate Stress Expressions
From the definitions above, we can equate the two expressions for stress:
\[ \frac{F}{\pi r^2} = Y \cdot \alpha \Delta T \]

### Step 5: Solve for Coefficient of Linear Expansion
Rearranging the equation gives us:
\[ \alpha = \frac{F}{\pi r^2 Y \Delta T} \]

### Step 6: Relate Coefficient of Linear Expansion to Coefficient of Volume Expansion
The relationship between the coefficient of linear expansion (\( \alpha \)), the coefficient of superficial expansion (\( \beta \)), and the coefficient of volume expansion (\( \gamma \)) is given by:
\[ \gamma = 3\alpha \]
Thus, substituting for \( \alpha \):
\[ \gamma = 3 \cdot \frac{F}{\pi r^2 Y \Delta T} \]

### Final Expression
Therefore, the coefficient of volume expansion \( \gamma \) is:
\[ \gamma = \frac{3F}{\pi r^2 Y \Delta T} \]