A glass cylinder contains `m_0=100g` of mercury at a temperature of `t_0=0^@C`. When temperature becomes `t_1=20^@C` the cylinder contains `m_1=99.7g` of mercury The coefficient of volume expansion of mercury `gamma_(He)=18xx(10^(-5)//^(@)C` Assume that the temperature of the mercury is equal to that of the cylinder. The coefficient of linear expansion of glass `alpha` is

Options

A. `=10^(-4)//.^(@)C`
B. `=10^(-3)//.^(@)C`
C. `=10^(-5)//.^(@)C`
D. `=10^(-2)//.^(@)C`

Answer

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

### Step 1: Understand the relationship between mass, volume, and density
The mass of mercury in the cylinder at the initial temperature \( t_0 \) is given by:
\[ m_0 = V_0 \cdot \rho_0 \]
where \( V_0 \) is the initial volume of mercury and \( \rho_0 \) is the initial density of mercury.

At a higher temperature \( t_1 \), the mass of mercury becomes \( m_1 \):
\[ m_1 = V_1 \cdot \rho_1 \]
where \( V_1 \) is the new volume of mercury and \( \rho_1 \) is the new density of mercury.

### Step 2: Relate the volumes and densities
We can express the density at temperature \( t_1 \) using the volume expansion formula:
\[ V_1 = V_0 \cdot (1 + \gamma_{Hg} \cdot (t_1 - t_0)) \]
Substituting this into the mass equation gives:
\[ m_1 = V_0 \cdot (1 + \gamma_{Hg} \cdot (t_1 - t_0)) \cdot \rho_1 \]

### Step 3: Substitute known values
Given:
- \( m_0 = 100 \, g \)
- \( m_1 = 99.7 \, g \)
- \( \gamma_{Hg} = 18 \times 10^{-5} \, ^\circ C^{-1} \)
- \( t_0 = 0 \, ^\circ C \)
- \( t_1 = 20 \, ^\circ C \)

Substituting these values into the equation:
\[ 99.7 = 100 \cdot (1 + 18 \times 10^{-5} \cdot (20 - 0)) \cdot \rho_1 \]

### Step 4: Calculate the new density
First, calculate the change in volume due to the expansion of mercury:
\[ 1 + \gamma_{Hg} \cdot (20 - 0) = 1 + 18 \times 10^{-5} \cdot 20 = 1 + 0.00036 = 1.00036 \]

Now substitute this back into the mass equation:
\[ 99.7 = 100 \cdot 1.00036 \cdot \rho_1 \]
Solving for \( \rho_1 \):
\[ \rho_1 = \frac{99.7}{100 \cdot 1.00036} \approx 0.9963 \, \rho_0 \]

### Step 5: Relate the change in density to the expansion of the glass
The density of the glass cylinder also changes due to thermal expansion:
\[ \rho_1 = \rho_0 \cdot \frac{1}{1 + \gamma_{g} \cdot (t_1 - t_0)} \]
Setting the two expressions for \( \rho_1 \) equal:
\[ 0.9963 \, \rho_0 = \rho_0 \cdot \frac{1}{1 + \gamma_{g} \cdot 20} \]

### Step 6: Solve for the coefficient of linear expansion of glass (α)
Cancelling \( \rho_0 \) from both sides:
\[ 0.9963 = \frac{1}{1 + \gamma_{g} \cdot 20} \]
Rearranging gives:
\[ 1 + \gamma_{g} \cdot 20 = \frac{1}{0.9963} \]
Calculating the right-hand side:
\[ 1 + \gamma_{g} \cdot 20 \approx 1.0037 \]
Thus:
\[ \gamma_{g} \cdot 20 \approx 0.0037 \]
\[ \gamma_{g} \approx \frac{0.0037}{20} = 1.85 \times 10^{-4} \, ^\circ C^{-1} \]

### Step 7: Convert volume expansion to linear expansion
Since the coefficient of linear expansion \( \alpha \) is related to the volume expansion \( \gamma \) by:
\[ \alpha = \frac{\gamma}{3} \]
Substituting the value of \( \gamma_{g} \):
\[ \alpha = \frac{1.85 \times 10^{-4}}{3} \approx 6.17 \times 10^{-5} \, ^\circ C^{-1} \]

### Final Answer
The coefficient of linear expansion of glass \( \alpha \) is approximately:
\[ \alpha \approx 6.17 \times 10^{-5} \, ^\circ C^{-1} \]