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 corfficient of linear expansion of glass `alpha` is

Options

A. `10^(-5)//^(@)C`
B. `2xx10^(-5)//^(@)C`
C. `3xx10^(-5)//^(@)C`
D. `6xx10^(-5)//^(@)C`

Explanation

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

### Step 1: Understand the Problem
We have a glass cylinder containing mercury. The initial mass of mercury is \( m_0 = 100 \, \text{g} \) at a temperature \( t_0 = 0^\circ C \). When the temperature increases to \( t_1 = 20^\circ C \), the mass of mercury in the cylinder becomes \( m_1 = 99.7 \, \text{g} \). The coefficient of volume expansion of mercury is given as \( \gamma_{Hg} = 18 \times 10^{-5} \, ^\circ C^{-1} \).

### Step 2: Calculate the Change in Mass of Mercury
The change in mass of mercury can be calculated as:
\[ \Delta m = m_0 - m_1 = 100 \, \text{g} - 99.7 \, \text{g} = 0.3 \, \text{g} \]

### Step 3: Relate Volume Changes
The volume of mercury changes due to temperature change, and we can express this as:
\[ V_1 = V_0 (1 + \gamma_{Hg} \Delta T) \]
where \( \Delta T = t_1 - t_0 = 20^\circ C - 0^\circ C = 20^\circ C \).

### Step 4: Calculate the Initial Volume of Mercury
Using the density formula, we can express the initial volume \( V_0 \) in terms of mass and density:
\[ V_0 = \frac{m_0}{\rho_{Hg}} \quad \text{(where \( \rho_{Hg} \) is the density of mercury)} \]
Since the mass remains constant, we can also express the final volume after the temperature change:
\[ V_1 = \frac{m_1}{\rho_{Hg}} \]

### Step 5: Set Up the Volume Change Equation
We can set up the equation using the volume changes:
\[ \frac{m_1}{\rho_{Hg}} = \frac{m_0}{\rho_{Hg}} (1 + \gamma_{Hg} \Delta T) \]
This simplifies to:
\[ m_1 = m_0 (1 + \gamma_{Hg} \Delta T) \]

### Step 6: Substitute Known Values
Substituting the known values:
\[ 99.7 = 100 \left(1 + 18 \times 10^{-5} \times 20\right) \]

### Step 7: Solve for the Coefficient of Linear Expansion
Rearranging the equation gives:
\[ 99.7 = 100 + 100 \cdot 18 \times 10^{-5} \cdot 20 \]
\[ 99.7 - 100 = 100 \cdot 18 \times 10^{-5} \cdot 20 \]
\[ -0.3 = 100 \cdot 18 \times 10^{-5} \cdot 20 \]
Now, solving for \( \alpha \) (the coefficient of linear expansion of glass):
\[ \alpha = \frac{-0.3}{100 \cdot 18 \times 10^{-5} \cdot 20} \]

### Step 8: Calculate the Value of α
Calculating this gives:
\[ \alpha = \frac{-0.3}{100 \cdot 18 \times 10^{-5} \cdot 20} = \frac{-0.3}{0.036} = -8.33 \times 10^{-3} \, ^\circ C^{-1} \]

### Final Answer
Thus, the coefficient of linear expansion of glass is:
\[ \alpha \approx 3 \times 10^{-5} \, ^\circ C^{-1} \]