If the density of the material increase, the value of young's modulus

Options

A. increases
B. decreases
C. first increases, then decreases
D. first decreases, then increases

Answer

To determine how the value of Young's modulus changes with an increase in the density of a material, we can follow these steps:

### Step 1: Understand Young's Modulus
Young's modulus (Y) is defined as the ratio of stress (force per unit area) to strain (change in length per original length) in a material. The formula for Young's modulus is given by:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
where:
- \( F \) = force applied,
- \( L \) = original length of the material,
- \( A \) = cross-sectional area,
- \( \Delta L \) = change in length.

### Step 2: Rearranging the Formula
We can manipulate the formula to express it in terms of volume. We know that:
\[ \text{Volume} (V) = A \cdot L \]
Thus, we can express the area \( A \) in terms of volume:
\[ A = \frac{V}{L} \]
Substituting this back into the Young's modulus equation gives:
\[ Y = \frac{F \cdot L}{\left(\frac{V}{L}\right) \cdot \Delta L} = \frac{F \cdot L^2}{V \cdot \Delta L} \]

### Step 3: Relate Volume to Density
Density (\( \rho \)) is defined as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass of the material. Rearranging gives:
\[ V = \frac{m}{\rho} \]
Substituting this expression for volume into the Young's modulus equation results in:
\[ Y = \frac{F \cdot L^2 \cdot \rho}{m \cdot \Delta L} \]

### Step 4: Analyze the Relationship
From the equation derived, we can see that Young's modulus is directly proportional to the density (\( \rho \)). This means that as the density of the material increases, the value of Young's modulus also increases, assuming that the force, length, and change in length remain constant.

### Conclusion
Therefore, if the density of the material increases, the value of Young's modulus also increases.

### Final Answer
The correct answer is: **Young's modulus increases with an increase in density.**

---