A uniform tube closed at one end, contains a pallet of mercury 4 cm long. When the tube is kept vertically with closed end upwards, the length of air column between closed end and mercury pallet is 10 cm. the tube is inverted so that open end becomes upwards. find the length of trapped air column (in cm) trapped. take atmospheric pressure 76 cm of Hg.

Answer

To solve the problem step by step, we will analyze the situation before and after the tube is inverted.

### Step 1: Understand the initial conditions
- The tube is closed at one end and contains a mercury pellet that is 4 cm long.
- When the tube is vertical with the closed end upwards, the length of the air column above the mercury pellet is 10 cm.
- The atmospheric pressure is given as 76 cm of Hg.

### Step 2: Calculate the pressure at the closed end when the tube is upright
Using the information provided:
- The total pressure at the closed end (P1) plus the height of the mercury column (4 cm) equals the atmospheric pressure (76 cm):
\[ P_1 + 4 \text{ cm} = 76 \text{ cm} \]
\[ P_1 = 76 \text{ cm} - 4 \text{ cm} = 72 \text{ cm} \]

### Step 3: Analyze the situation after inverting the tube
When the tube is inverted:
- The closed end is now at the bottom, and the open end is at the top.
- The mercury pellet remains 4 cm long, but now it is at the bottom of the tube.

### Step 4: Calculate the pressure at the open end when inverted
The pressure at the open end (P2) now includes the atmospheric pressure and the height of the mercury column:
\[ P_2 = P_{\text{atm}} + \text{height of mercury column} \]
Here, the height of the mercury column is still 4 cm:
\[ P_2 = 76 \text{ cm} + 4 \text{ cm} = 80 \text{ cm} \]

### Step 5: Use the ideal gas law to relate the two states
According to the ideal gas law, the product of pressure and volume remains constant for a given amount of gas:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( V_1 \) is the volume of the air column when the tube is upright.
- \( V_2 \) is the volume of the air column when the tube is inverted.

### Step 6: Calculate the volume of the air column
The volume of the air column can be expressed as:
\[ V_1 = A \cdot h_1 = A \cdot 10 \text{ cm} \]
\[ V_2 = A \cdot h_2 \]
Where \( h_1 = 10 \text{ cm} \) and \( h_2 \) is the unknown length of the air column when inverted.

### Step 7: Substitute into the ideal gas equation
Substituting the volumes into the equation:
\[ P_1 (A \cdot 10) = P_2 (A \cdot h_2) \]
Canceling the area \( A \) from both sides:
\[ P_1 \cdot 10 = P_2 \cdot h_2 \]
Substituting the values of \( P_1 \) and \( P_2 \):
\[ 72 \cdot 10 = 80 \cdot h_2 \]
\[ 720 = 80 h_2 \]
\[ h_2 = \frac{720}{80} = 9 \text{ cm} \]

### Final Answer
The length of the trapped air column when the tube is inverted is **9 cm**.

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