The temperature of a metal ball is raised. Arrange the percentage change in volume, surface area and raidus in ascending order.

Answer

To solve the problem of arranging the percentage change in volume, surface area, and radius of a metal ball when its temperature is raised, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Coefficients of Expansion**:
- The coefficient of linear expansion (α) describes how the length of an object changes with temperature.
- The coefficient of superficial (area) expansion (β) is related to the linear expansion as follows: \( \beta = 2\alpha \).
- The coefficient of volume expansion (γ) is related to the linear expansion as follows: \( \gamma = 3\alpha \).

2. **Express Percentage Changes**:
- The percentage change in volume (ΔV/V) can be expressed as:
\[ \frac{\Delta V}{V} = \gamma \Delta T = 3\alpha \Delta T \]
- The percentage change in surface area (ΔA/A) can be expressed as:
\[ \frac{\Delta A}{A} = \beta \Delta T = 2\alpha \Delta T \]
- The percentage change in radius (ΔR/R) can be expressed as:
\[ \frac{\Delta R}{R} = \alpha \Delta T \]

3. **Compare the Percentage Changes**:
- Now we can compare the three expressions:
- Percentage change in radius: \( \frac{\Delta R}{R} = \alpha \Delta T \)
- Percentage change in area: \( \frac{\Delta A}{A} = 2\alpha \Delta T \)
- Percentage change in volume: \( \frac{\Delta V}{V} = 3\alpha \Delta T \)

4. **Arrange in Ascending Order**:
- From the comparisons:
- \( \frac{\Delta R}{R} < \frac{\Delta A}{A} < \frac{\Delta V}{V} \)
- Therefore, the ascending order of percentage changes is:
\[ \text{Percentage change in radius} < \text{Percentage change in area} < \text{Percentage change in volume} \]

### Final Answer:
The ascending order of percentage change is:
- Percentage change in radius < Percentage change in area < Percentage change in volume