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

Explanation

To find the coefficient of linear expansion (γ) of the substance given its densities at two temperatures, 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):
\[ \rho = \frac{m}{V} \]
When the temperature changes, the volume of the substance changes due to thermal expansion. The new volume (V) at a higher temperature can be expressed as:
\[ V = V_0(1 + \gamma \Delta T) \]
where:
- \(V_0\) is the initial volume,
- \(\gamma\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 2: Set up the equations for the two temperatures
Given:
- At \(0^\circ C\), \(\rho_1 = 10 \, \text{g/cc}\)
- At \(100^\circ C\), \(\rho_2 = 9.7 \, \text{g/cc}\)

From the definitions of density, we can write:
\[ \rho_1 = \frac{m}{V_0} \quad \text{and} \quad \rho_2 = \frac{m}{V} \]
Substituting for \(V\) using the volume expansion formula:
\[ \rho_2 = \frac{m}{V_0(1 + \gamma \Delta T)} \]
where \(\Delta T = 100 - 0 = 100\).

### Step 3: Relate the two densities
Now we can write:
\[ \rho_2 = \frac{\rho_1 V_0}{V_0(1 + \gamma \cdot 100)} \]
This simplifies to:
\[ \rho_2 = \frac{\rho_1}{1 + \gamma \cdot 100} \]

### Step 4: Substitute the known values
Substituting the values of \(\rho_1\) and \(\rho_2\):
\[ 9.7 = \frac{10}{1 + 100\gamma} \]

### Step 5: Cross-multiply to solve for \(\gamma\)
Cross-multiplying gives:
\[ 9.7(1 + 100\gamma) = 10 \]
Expanding this:
\[ 9.7 + 970\gamma = 10 \]

### Step 6: Rearrange to isolate \(\gamma\)
Rearranging gives:
\[ 970\gamma = 10 - 9.7 \]
\[ 970\gamma = 0.3 \]
\[ \gamma = \frac{0.3}{970} \]

### Step 7: Calculate \(\gamma\)
Calculating this gives:
\[ \gamma \approx 3.09 \times 10^{-4} \, \text{°C}^{-1} \]

### Final Answer
The coefficient of linear expansion of the substance is approximately:
\[ \gamma \approx 3 \times 10^{-4} \, \text{°C}^{-1} \]