A metallic ball has spherical cavity at its centre. If the ball is heated, what happens to the cavity

Options

A. Its volume decreases.
B. Its volume remains constant.
C. Its volume increases.
D. Nothing can be decided.

Explanation

To solve the question about what happens to a spherical cavity at the center of a metallic ball when the ball is heated, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- We have a metallic ball with a spherical cavity at its center. Let's denote the radius of the metallic ball as \( R \) and the radius of the cavity as \( r \).

2. **Effect of Heating**:
- When the metallic ball is heated, the entire material of the ball expands due to thermal expansion. This is a property of metals where they tend to increase in volume when their temperature rises.

3. **Volume Expansion**:
- The volume expansion of the metallic ball can be described by the formula:
\[ V = V_0 (1 + \gamma \Delta T) \]
where \( V_0 \) is the initial volume, \( \gamma \) is the coefficient of volume expansion, and \( \Delta T \) is the change in temperature.

4. **Calculating Initial Volume**:
- The initial volume of the metallic ball (excluding the cavity) can be expressed as:
\[ V_0 = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (R^3 - r^3) \]

5. **Volume Increase**:
- As the temperature increases, the volume \( V \) of the metallic ball will increase. Since the volume of the ball is directly related to its radius (specifically, the radius cubed), both the radius of the metallic ball and the radius of the cavity will increase.

6. **Conclusion**:
- Therefore, when the metallic ball is heated, both the metallic material surrounding the cavity and the cavity itself will expand. This means that the radius of the cavity will increase as well.

### Final Answer:
When the metallic ball is heated, the cavity at its center also expands, resulting in an increase in the radius of the cavity.

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