The coefficient of apparent expansion of mercury in a glass vessel is `153xx10^(-6)//^(@)C` and in a steel vessel is `114xx10^(-6)//^(@)C`. If `alpha` for steel is `12xx10^(-6)//^(@)C`, then that of glass is

Options

A. `9xx10^(-6)//^(@)C`
B. `6xx10^(-6)//^(@)C`
C. `36xx10^(-6)//^(@)C`
D. `27xx(10^(-6)//^(@)C`

Explanation

To solve the problem, we need to find the coefficient of linear expansion (α) for glass given the apparent expansion coefficients for mercury in glass and steel vessels, and the coefficient of linear expansion for steel.

### Step-by-Step Solution:

1. **Understanding the Apparent Expansion**:
The apparent coefficient of expansion (γ) of a liquid in a vessel is given by the formula:
\[ \gamma = \gamma_{\text{liquid}} - \gamma_{\text{vessel}} \]
where:
- \(\gamma_{\text{liquid}}\) is the volumetric expansion coefficient of the liquid.
- \(\gamma_{\text{vessel}}\) is the volumetric expansion coefficient of the vessel.

2. **Given Values**:
- Apparent expansion of mercury in glass: \(\gamma_{\text{apparent, glass}} = 153 \times 10^{-6} \, \text{°C}^{-1}\)
- Apparent expansion of mercury in steel: \(\gamma_{\text{apparent, steel}} = 144 \times 10^{-6} \, \text{°C}^{-1}\)
- Coefficient of volumetric expansion for steel: \(\alpha_{\text{steel}} = 12 \times 10^{-6} \, \text{°C}^{-1}\)

3. **Relating Volumetric Expansion to Linear Expansion**:
The volumetric expansion coefficient (γ) is related to the linear expansion coefficient (α) by:
\[ \gamma = 3\alpha \]
Thus, for steel:
\[ \gamma_{\text{steel}} = 3\alpha_{\text{steel}} = 3 \times (12 \times 10^{-6}) = 36 \times 10^{-6} \, \text{°C}^{-1} \]

4. **Setting Up the Equations**:
For the glass vessel:
\[ \gamma_{\text{mercury}} = \gamma_{\text{apparent, glass}} + \gamma_{\text{glass}} \]
For the steel vessel:
\[ \gamma_{\text{mercury}} = \gamma_{\text{apparent, steel}} + \gamma_{\text{steel}} \]

5. **Equating the Two Expressions**:
Since both expressions equal \(\gamma_{\text{mercury}}\), we can set them equal to each other:
\[ \gamma_{\text{apparent, glass}} + \gamma_{\text{glass}} = \gamma_{\text{apparent, steel}} + \gamma_{\text{steel}} \]

6. **Substituting Known Values**:
Substitute the known values into the equation:
\[ 153 \times 10^{-6} + \gamma_{\text{glass}} = 144 \times 10^{-6} + 36 \times 10^{-6} \]

7. **Simplifying the Equation**:
Simplifying the right side:
\[ 144 \times 10^{-6} + 36 \times 10^{-6} = 180 \times 10^{-6} \]
Now we have:
\[ 153 \times 10^{-6} + \gamma_{\text{glass}} = 180 \times 10^{-6} \]

8. **Solving for \(\gamma_{\text{glass}}\)**:
Rearranging gives:
\[ \gamma_{\text{glass}} = 180 \times 10^{-6} - 153 \times 10^{-6} = 27 \times 10^{-6} \]

9. **Finding \(\alpha_{\text{glass}}\)**:
Since \(\gamma_{\text{glass}} = 3\alpha_{\text{glass}}\):
\[ 3\alpha_{\text{glass}} = 27 \times 10^{-6} \]
Therefore:
\[ \alpha_{\text{glass}} = \frac{27 \times 10^{-6}}{3} = 9 \times 10^{-6} \, \text{°C}^{-1} \]

### Final Answer:
The coefficient of linear expansion for glass is:
\[ \alpha_{\text{glass}} = 9 \times 10^{-6} \, \text{°C}^{-1} \]