To solve the problem step by step, we will break it down into two parts as given in the question.
### Part (a)
1. **Identify the given values**:
- Coefficient of linear expansion of brass, \( \alpha = 2.0 \times 10^{-5} \, ^\circ C^{-1} \)
- Initial temperature, \( T_1 = 0^\circ C \)
- Final temperature, \( T_2 = 27^\circ C \)
- Barometer reading at \( T_2 \), \( L + \Delta L = 75 \, cm \)
2. **Calculate the change in temperature**:
\[
\Delta T = T_2 - T_1 = 27^\circ C - 0^\circ C = 27^\circ C
\]
3. **Calculate the change in length due to thermal expansion**:
\[
\Delta L = \alpha L \Delta T
\]
Since we need to find \( L \), we can express it as:
\[
L + \Delta L = 75 \, cm \implies L + \alpha L \Delta T = 75 \, cm
\]
4. **Factor out \( L \)**:
\[
L(1 + \alpha \Delta T) = 75 \, cm
\]
5. **Solve for \( L \)**:
\[
L = \frac{75 \, cm}{1 + \alpha \Delta T}
\]
Substituting the values:
\[
L = \frac{75 \, cm}{1 + (2.0 \times 10^{-5}) \times 27}
\]
\[
L = \frac{75 \, cm}{1 + 0.00054} = \frac{75 \, cm}{1.00054} \approx 74.95 \, cm
\]
6. **Convert the length to atmospheric pressure**:
Atmospheric pressure \( P \) in atm is given by:
\[
P = \frac{L \times 100}{760}
\]
Thus,
\[
P = \frac{74.95 \times 100}{760} \approx 0.9861 \, atm
\]
### Part (b)
1. **Identify the given values**:
- Coefficient of linear expansion of steel, \( \alpha = 1.2 \times 10^{-5} \, ^\circ C^{-1} \)
- Coefficient of volume expansion of mercury, \( \gamma = 1.8 \times 10^{-4} \, ^\circ C^{-1} \)
- Initial temperature, \( T_1 = 0^\circ C \)
- Final temperature, \( T_2 = 30^\circ C \)
- Barometer reading at \( T_2 \), \( L + \Delta L = 75.0 \, cm \)
2. **Calculate the change in temperature**:
\[
\Delta T = T_2 - T_1 = 30^\circ C - 0^\circ C = 30^\circ C
\]
3. **Calculate the change in length due to thermal expansion**:
\[
\Delta L = \alpha L \Delta T
\]
Again, we express it as:
\[
L + \Delta L = 75.0 \, cm \implies L + \alpha L \Delta T = 75.0 \, cm
\]
4. **Factor out \( L \)**:
\[
L(1 + \alpha \Delta T) = 75.0 \, cm
\]
5. **Solve for \( L \)**:
\[
L = \frac{75.0 \, cm}{1 + \alpha \Delta T}
\]
Substituting the values:
\[
L = \frac{75.0 \, cm}{1 + (1.2 \times 10^{-5}) \times 30}
\]
\[
L = \frac{75.0 \, cm}{1 + 0.00036} = \frac{75.0 \, cm}{1.00036} \approx 74.973 \, cm
\]
6. **Calculate the change in volume of mercury**:
The change in height of mercury due to volume expansion is given by:
\[
L_1 + \Delta L_1 = L + \Delta L_1
\]
Where \( \Delta L_1 = \frac{\gamma}{3} L_1 \Delta T \).
7. **Express the equation**:
\[
L_1(1 + \frac{\gamma}{3} \Delta T) = 74.973 \, cm
\]
8. **Solve for \( L_1 \)**:
\[
L_1 = \frac{74.973 \, cm}{1 + \frac{1.8 \times 10^{-4}}{3} \times 30}
\]
\[
L_1 = \frac{74.973 \, cm}{1 + 0.0018} = \frac{74.973 \, cm}{1.0018} \approx 74.838 \, cm
\]
9. **Convert the length to atmospheric pressure**:
\[
P = \frac{L_1 \times 100}{760}
\]
Thus,
\[
P = \frac{74.838 \times 100}{760} \approx 0.98478 \, atm
\]
