A glass flask whose volume is exactly `1000cm^3` at `0^@C` is filled level full of mercury at this temperature. When the flask and mercury are heated to `100^@C`, `15.2cm^3` of mercury overflows. The coefficient of cubical expansion of Hg is `1.82xx10^(-4)//^@ C`. Compute the coefficient of linear expansion of glass.

Options

A. `0.1xx10^(-5)//^@C`
B. `1xx10^(-4)//^@C`
C. `0.1xx10^(-4)//^@C`
D. `1xx10^(-5)//^@C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a glass flask with a volume of 1000 cm³ at 0°C filled with mercury. When heated to 100°C, 15.2 cm³ of mercury overflows. We need to find the coefficient of linear expansion of glass.

### Step 2: Write the Overflow Equation
The overflow of the liquid (mercury) is given by the equation:
\[ \text{Overflow} = V_{\text{final, liquid}} - V_{\text{final, container}} \]

### Step 3: Calculate the Final Volume of Mercury
The final volume of mercury when heated can be expressed using the formula for cubical expansion:
\[ V_{\text{final, liquid}} = V_0 \times (1 + \gamma_{\text{Hg}} \times \Delta T) \]
Where:
- \( V_0 = 1000 \, \text{cm}^3 \)
- \( \gamma_{\text{Hg}} = 1.82 \times 10^{-4} \, /^{\circ}C \)
- \( \Delta T = 100^{\circ}C \)

Substituting the values:
\[ V_{\text{final, liquid}} = 1000 \times (1 + 1.82 \times 10^{-4} \times 100) \]
\[ V_{\text{final, liquid}} = 1000 \times (1 + 0.0182) \]
\[ V_{\text{final, liquid}} = 1000 \times 1.0182 = 1018.2 \, \text{cm}^3 \]

### Step 4: Calculate the Final Volume of the Flask
The final volume of the glass flask can be expressed similarly:
\[ V_{\text{final, container}} = V_0 \times (1 + \gamma_{\text{glass}} \times \Delta T) \]
Where \( \gamma_{\text{glass}} \) is what we need to find.

Substituting the values:
\[ V_{\text{final, container}} = 1000 \times (1 + \gamma_{\text{glass}} \times 100) \]

### Step 5: Set Up the Equation for Overflow
From the overflow equation:
\[ 15.2 = V_{\text{final, liquid}} - V_{\text{final, container}} \]
Substituting the expressions we derived:
\[ 15.2 = 1018.2 - (1000 \times (1 + \gamma_{\text{glass}} \times 100)) \]

### Step 6: Simplify the Equation
Rearranging gives:
\[ 15.2 = 1018.2 - 1000 - 1000 \gamma_{\text{glass}} \times 100 \]
\[ 15.2 = 18.2 - 100000 \gamma_{\text{glass}} \]
\[ 100000 \gamma_{\text{glass}} = 18.2 - 15.2 \]
\[ 100000 \gamma_{\text{glass}} = 3 \]
\[ \gamma_{\text{glass}} = \frac{3}{100000} = 3 \times 10^{-5} \, /^{\circ}C \]

### Step 7: Relate Cubical Expansion to Linear Expansion
The relationship between the cubical expansion coefficient (\( \gamma \)) and the linear expansion coefficient (\( \alpha \)) is given by:
\[ \gamma = 3\alpha \]
Thus,
\[ \alpha = \frac{\gamma_{\text{glass}}}{3} = \frac{3 \times 10^{-5}}{3} = 1 \times 10^{-5} \, /^{\circ}C \]

### Final Answer
The coefficient of linear expansion of glass is:
\[ \alpha_{\text{glass}} = 1 \times 10^{-5} \, /^{\circ}C \]

---