The density of a substance at `0^(@)C` is `10 g//c c` and at `100^(@)C`, its density is `9.7 g//c c`. The coefficient of linear expansion of the substance is

Options

A. `10^(2)`
B. `10^(-2)`
C. `10^(-3)`
D. `10^(-4)`

Explanation

To find the coefficient of linear expansion (α) of the substance, we can follow these steps:

### Step 1: Understand the relationship between density and volume
The density (ρ) of a substance is defined as its mass (m) divided by its volume (V). Therefore, we can express the density at two different temperatures as follows:
- At 0°C: \( \rho_0 = \frac{m}{V_0} \)
- At 100°C: \( \rho = \frac{m}{V} \)

### Step 2: Relate the volumes at different temperatures
When a substance is heated, it expands, and its volume changes. The relationship between the original volume (V₀) and the new volume (V) after a temperature change (ΔT) is given by:
\[ V = V_0(1 + \gamma \Delta T) \]
where γ is the coefficient of volumetric expansion.

### Step 3: Substitute the volume relationship into the density equation
We can express the new density (ρ) in terms of the original density (ρ₀) and the coefficient of volumetric expansion (γ):
\[ \rho = \frac{m}{V} = \frac{m}{V_0(1 + \gamma \Delta T)} = \frac{\rho_0}{1 + \gamma \Delta T} \]

### Step 4: Calculate the change in density
From the problem, we know:
- \( \rho_0 = 10 \, g/cm^3 \) (at 0°C)
- \( \rho = 9.7 \, g/cm^3 \) (at 100°C)

The change in density (Δρ) is:
\[ \Delta \rho = \rho_0 - \rho = 10 - 9.7 = 0.3 \, g/cm^3 \]

### Step 5: Use the density relationship to find γ
From the density relationship, we can rearrange it to find γ:
\[ \Delta \rho = \rho \cdot \gamma \Delta T \]
Substituting the known values:
\[ 0.3 = 9.7 \cdot \gamma \cdot 100 \]

### Step 6: Solve for γ
Rearranging the equation gives:
\[ \gamma = \frac{0.3}{9.7 \cdot 100} \]
\[ \gamma = \frac{0.3}{970} \]
\[ \gamma = 3.09 \times 10^{-4} \, °C^{-1} \]

### Step 7: Calculate the coefficient of linear expansion (α)
The relationship between the coefficient of volumetric expansion (γ) and the coefficient of linear expansion (α) is given by:
\[ \alpha = \frac{\gamma}{3} \]
Substituting the value of γ:
\[ \alpha = \frac{3.09 \times 10^{-4}}{3} \]
\[ \alpha = 1.03 \times 10^{-4} \, °C^{-1} \]

### Final Answer
The coefficient of linear expansion (α) of the substance is approximately \( 1.03 \times 10^{-4} \, °C^{-1} \).

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