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
When the tube is vertical with the closed end upwards:
- The length of the mercury column is 4 cm.
- The length of the air column above the mercury is 10 cm.
- The total height of the mercury column and the air column is 14 cm.

### Step 2: Calculate the Pressure of the Air Column (P1)
Using the atmospheric pressure (P_atm) given as 76 cm of Hg:
- The pressure exerted by the air column (P1) plus the pressure exerted by the mercury column (4 cm of Hg) must equal the atmospheric pressure.

\[ P1 + 4 \text{ cm} = 76 \text{ cm} \]

Rearranging gives:

\[ P1 = 76 \text{ cm} - 4 \text{ cm} = 72 \text{ cm} \]

### Step 3: Invert the Tube
When the tube is inverted:
- The closed end is now downwards, and the open end is upwards.
- The mercury column remains 4 cm long, but now it will exert pressure downwards.

### Step 4: Calculate the Pressure of the Air Column After Inversion (P2)
The pressure exerted by the air column (P2) now must balance the atmospheric pressure plus the pressure due to the mercury column:

\[ P2 = P_{atm} + 4 \text{ cm} \]

Substituting the atmospheric pressure:

\[ P2 = 76 \text{ cm} + 4 \text{ cm} = 80 \text{ cm} \]

### Step 5: Relate the Volumes Before and After Inversion
Using the ideal gas law, we know that:

\[ P1 \cdot V1 = P2 \cdot V2 \]

Where:
- \( V1 \) is the volume of the air column before inversion.
- \( V2 \) is the volume of the air column after inversion.

The volume of the air column can be expressed as:
- \( V1 = A \cdot h_1 = A \cdot 10 \text{ cm} \) (where A is the cross-sectional area)
- \( V2 = A \cdot h_2 = A \cdot D \) (where D is the new height of the air column we want to find)

### Step 6: Substitute into the Equation
Substituting the volumes into the equation gives:

\[ 72 \cdot (10A) = 80 \cdot (DA) \]

### Step 7: Cancel the Area
Since A is present in both sides, we can cancel it out:

\[ 72 \cdot 10 = 80 \cdot D \]

### Step 8: Solve for D
Now, we can solve for D:

\[ 720 = 80D \]

Dividing both sides by 80:

\[ D = \frac{720}{80} = 9 \text{ cm} \]

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

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