A barometer having a brass scale reads 77.24 cm at a temperature of `20^(@)C`. The scale is graduated to be accurate at `0^(@)C`. What would be the readint at `0^(@)C` ? Coefficient of cubical expansion of mercury `=18xx10^(-5).^(@)C^(-1)` and linear expansion of brass `-19xx10^(-6).^(@)C^(-1)`

Answer

To solve the problem, we need to determine the reading of the barometer at 0°C, given that it reads 77.24 cm at 20°C. The scale is calibrated to be accurate at 0°C, and we have the coefficients of thermal expansion for both mercury and brass.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Reading at 20°C, \( L_{20} = 77.24 \) cm
- Temperature at which the reading is taken, \( T_1 = 20°C \)
- Temperature at which we want to find the reading, \( T_2 = 0°C \)
- Coefficient of cubical expansion of mercury, \( \beta_{Hg} = 18 \times 10^{-5} \, °C^{-1} \)
- Coefficient of linear expansion of brass, \( \alpha_{brass} = 19 \times 10^{-6} \, °C^{-1} \)

2. **Calculate the Change in Volume of Mercury:**
The change in volume of mercury due to temperature change can be calculated using the formula:
\[ \Delta V_{Hg} = V_{initial} \cdot \beta_{Hg} \cdot \Delta T \]
Since we are dealing with a linear measurement (height of mercury), we can relate the change in height to the change in volume:
\[ \Delta L_{Hg} = L_{20} \cdot \beta_{Hg} \cdot (T_1 - T_2) \]
Substituting the values:
\[ \Delta L_{Hg} = 77.24 \cdot 18 \times 10^{-5} \cdot (20 - 0) \]
\[ \Delta L_{Hg} = 77.24 \cdot 18 \times 10^{-5} \cdot 20 \]
\[ \Delta L_{Hg} = 77.24 \cdot 0.00036 \approx 0.0278 \, \text{cm} \]

3. **Calculate the Change in Length of the Brass Scale:**
The change in length of the brass scale can be calculated using the formula:
\[ \Delta L_{brass} = L_{20} \cdot \alpha_{brass} \cdot (T_1 - T_2) \]
Substituting the values:
\[ \Delta L_{brass} = 77.24 \cdot 19 \times 10^{-6} \cdot (20 - 0) \]
\[ \Delta L_{brass} = 77.24 \cdot 19 \times 10^{-6} \cdot 20 \]
\[ \Delta L_{brass} = 77.24 \cdot 0.00038 \approx 0.0293 \, \text{cm} \]

4. **Calculate the Effective Reading at 0°C:**
The effective reading at 0°C can be found by adjusting the original reading at 20°C for the changes in both the mercury and the brass scale:
\[ L_{0} = L_{20} - \Delta L_{Hg} + \Delta L_{brass} \]
Substituting the values:
\[ L_{0} = 77.24 - 0.0278 + 0.0293 \]
\[ L_{0} = 77.24 + 0.0015 \approx 77.2415 \, \text{cm} \]

5. **Final Answer:**
The reading of the barometer at 0°C is approximately **77.24 cm**.