The brass scale of a barometer gives correct reading at `10^(@)C`. The baometer reads `75 cm` at `30^(@)C`. What is the a atmoshpheric pressure at `o^(@)C ("in cm "Hg)`
`(alpha_("brass")=20xx10^(-6)//^(0)C,lambda_(Hg)=175xx10^(-6)//^(@)C)`

Options

A. `74.8`
B. `75.03`
C. `70`
D. `60`

Explanation

To solve the problem, we need to find the atmospheric pressure at \(0^\circ C\) using the readings from the barometer at \(10^\circ C\) and \(30^\circ C\). We'll follow these steps:

### Step 1: Understand the given data
- The barometer gives a correct reading at \(10^\circ C\).
- The barometer reads \(75 \, \text{cm}\) at \(30^\circ C\).
- The coefficient of linear expansion for brass, \(\alpha_{\text{brass}} = 20 \times 10^{-6} \, ^\circ C^{-1}\).
- The coefficient of linear expansion for mercury, \(\lambda_{\text{Hg}} = 175 \times 10^{-6} \, ^\circ C^{-1}\).

### Step 2: Calculate the change in temperature
The change in temperature from \(30^\circ C\) to \(10^\circ C\) is:
\[ \Delta T = 10 - 30 = -20^\circ C \]

### Step 3: Calculate the effective coefficient of expansion
The effective change in reading due to the expansion of mercury and contraction of brass can be calculated using the formula:
\[ \text{Net coefficient} = \lambda_{\text{Hg}} - \alpha_{\text{brass}} = 175 \times 10^{-6} - 20 \times 10^{-6} = 155 \times 10^{-6} \, ^\circ C^{-1} \]

### Step 4: Calculate the change in reading
Using the effective coefficient, we can find the change in reading (\(\Delta L\)) of the barometer:
\[ \Delta L = L_0 \cdot \text{Net coefficient} \cdot \Delta T \]
Where \(L_0 = 75 \, \text{cm}\):
\[ \Delta L = 75 \cdot (155 \times 10^{-6}) \cdot (-20) \]
Calculating this gives:
\[ \Delta L = 75 \cdot 155 \cdot (-20) \times 10^{-6} = -0.2325 \, \text{cm} \]

### Step 5: Calculate the reading at \(10^\circ C\)
Now, we can find the reading at \(10^\circ C\):
\[ L_{10} = L_{30} + \Delta L = 75 - 0.2325 = 74.7675 \, \text{cm} \]

### Step 6: Round off the result
Rounding off \(74.7675 \, \text{cm}\) gives:
\[ L_{10} \approx 74.8 \, \text{cm} \]

### Conclusion
The atmospheric pressure at \(0^\circ C\) is approximately \(74.8 \, \text{cm}\) of mercury.

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