The volume of a solid decreases by 0.6% when it is cooled through `50^(@)C`. Its coefficient of linear expansion is

Options

A. `4xx10^(-6) K`
B. `5xx10^(-5) K`
C. `6xx10^(4)K`
D. `4xx10^(-5) K`

Explanation

To solve the problem, we need to find the coefficient of linear expansion (α) given that the volume of a solid decreases by 0.6% when it is cooled through 50°C.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the volume decreases by 0.6% when cooled through 50°C.
- We need to find the coefficient of linear expansion (α).

2. **Using the Formula for Volume Change**:
- The percentage change in volume (ΔV/V) can be expressed as:
\[ \frac{\Delta V}{V} \times 100 = \text{Percentage Change} \]
- Here, ΔV is the change in volume, and V is the initial volume.

3. **Relating Volume Change to Temperature Change**:
- The change in volume (ΔV) due to temperature change (ΔT) is given by:
\[ \Delta V = V \cdot \gamma \cdot \Delta T \]
- Where γ is the coefficient of volumetric expansion.

4. **Substituting into the Percentage Change Formula**:
- We can substitute ΔV in the percentage change formula:
\[ \frac{V \cdot \gamma \cdot \Delta T}{V} \times 100 = 0.6 \]
- This simplifies to:
\[ \gamma \cdot \Delta T \times 100 = 0.6 \]

5. **Finding the Coefficient of Volumetric Expansion**:
- Given that ΔT = 50°C, we can substitute this value:
\[ \gamma \cdot 50 \times 100 = 0.6 \]
- Rearranging gives:
\[ \gamma = \frac{0.6}{5000} = 1.2 \times 10^{-4} \]

6. **Relating Coefficients of Expansion**:
- The relationship between the coefficients of linear (α) and volumetric expansion (γ) is:
\[ \gamma = 3\alpha \]
- Therefore, we can express α in terms of γ:
\[ \alpha = \frac{\gamma}{3} = \frac{1.2 \times 10^{-4}}{3} = 0.4 \times 10^{-4} \]

7. **Final Calculation**:
- Simplifying gives:
\[ \alpha = 4 \times 10^{-5} \]

### Final Answer:
The coefficient of linear expansion (α) is:
\[ \alpha = 4 \times 10^{-5} \, \text{°C}^{-1} \]