When a copper ball is heated, the largest percentage increase will occur in its

Options

A. Diameter
B. Area
C. Volume
D. Density

Answer

To determine which property of a copper ball experiences the largest percentage increase when heated, we will analyze the changes in diameter, area, volume, and density due to thermal expansion.

### Step-by-Step Solution:

1. **Understanding Thermal Expansion**:
- When a material is heated, it expands. This expansion can be categorized into three types: linear expansion (change in length), area expansion (change in area), and volume expansion (change in volume).

2. **Linear Expansion (Diameter)**:
- The formula for linear expansion is given by:
\[ \Delta D = D \cdot \alpha \cdot \Delta T \]
- The percentage change in diameter is:
\[ \text{Percentage change in diameter} = \frac{\Delta D}{D} \times 100 = \alpha \cdot \Delta T \times 100 \]

3. **Area Expansion**:
- The area expansion is related to the linear expansion coefficient by:
\[ \beta = 2\alpha \]
- The percentage change in area is:
\[ \text{Percentage change in area} = \frac{\Delta A}{A} \times 100 = \beta \cdot \Delta T \times 100 = 2\alpha \cdot \Delta T \times 100 \]

4. **Volume Expansion**:
- The volume expansion is related to the linear expansion coefficient by:
\[ \gamma = 3\alpha \]
- The percentage change in volume is:
\[ \text{Percentage change in volume} = \frac{\Delta V}{V} \times 100 = \gamma \cdot \Delta T \times 100 = 3\alpha \cdot \Delta T \times 100 \]

5. **Density Change**:
- Density (\( \rho \)) is defined as mass (\( m \)) divided by volume (\( V \)):
\[ \rho = \frac{m}{V} \]
- When the volume increases due to heating, the mass remains constant, leading to a decrease in density. Thus, the percentage change in density is negative, indicating a decrease:
\[ \text{Percentage change in density} = \frac{\Delta \rho}{\rho} \times 100 < 0 \]

6. **Comparison of Percentage Changes**:
- From the above calculations:
- Percentage change in diameter: \( \alpha \cdot \Delta T \times 100 \)
- Percentage change in area: \( 2\alpha \cdot \Delta T \times 100 \)
- Percentage change in volume: \( 3\alpha \cdot \Delta T \times 100 \)
- Percentage change in density: Negative value (decrease)
- It is clear that the percentage change in volume is the largest among the three (diameter, area, and volume).

### Conclusion:
The largest percentage increase occurs in the **volume** of the copper ball when it is heated.