A quill tube contains a mercury column of length `19cm`. The length of air column is `24cm` when it is held veritcally. On inverting it with its open end downwards the length of air column will be
(atmospheric pressure `= 76cm` of `Hg`)

Options

A. `20cm`
B. `30cm`
C. `40cm`
D. `35cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
The quill tube has:
- A mercury column of length \( h_m = 19 \, \text{cm} \)
- An air column of length \( h_a = 24 \, \text{cm} \)
- Atmospheric pressure \( P_0 = 76 \, \text{cm of Hg} \)

### Step 2: Calculate the initial pressure of the air column
The pressure at the bottom of the air column when the tube is vertical can be expressed as:
\[ P = P_0 - \rho_{Hg} \cdot g \cdot h_m \]
Where \( \rho_{Hg} \) is the density of mercury and \( g \) is the acceleration due to gravity. However, since we are working with heights of mercury, we can express the pressure in terms of height of mercury directly.

Using the values:
\[ P = 76 \, \text{cm} - 19 \, \text{cm} = 57 \, \text{cm of Hg} \]

### Step 3: Invert the tube
When the tube is inverted, the mercury column will still be 19 cm, but the air column will change. Let the new length of the air column be \( L_1 \).

### Step 4: Calculate the pressure of the air column after inversion
After inversion, the pressure of the air column can be expressed as:
\[ P' = P_0 - \rho_{Hg} \cdot g \cdot h_m \]
This gives us:
\[ P' = 76 \, \text{cm} - 19 \, \text{cm} = 57 \, \text{cm of Hg} \]

### Step 5: Apply the ideal gas law (isothermal process)
Since the temperature remains constant during the process, we can use Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_1 = 95 \, \text{cm of Hg} \) (initial pressure of air)
- \( V_1 = 24 \, \text{cm} \) (initial volume of air)
- \( P_2 = 57 \, \text{cm of Hg} \) (pressure after inversion)
- \( V_2 = L_1 \) (new volume of air)

### Step 6: Substitute and solve for \( L_1 \)
Substituting the known values into the equation:
\[ 95 \times 24 = 57 \times L_1 \]
Now, solving for \( L_1 \):
\[ L_1 = \frac{95 \times 24}{57} \]

### Step 7: Calculate \( L_1 \)
Calculating:
\[ L_1 = \frac{2280}{57} \approx 40 \, \text{cm} \]

### Conclusion
Thus, the length of the air column after inverting the tube is \( L_1 = 40 \, \text{cm} \).

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