The lower end of a glass capillary tube is dipped in water. Water rises to a height of 9 cm. The tube is then broken at a height of 5 cm. The height of the water column and angle of contact will be

Options

A. `5 cm , cos ^(-1) (5/9)`
B. `4 cm , cos ^(-1) (5/4)`
C. `5 cm , cos ^(-1) (9/5)`
D. `5 cm , cos ^(-1) (6/7)`

Answer

To solve the problem step by step, we will analyze the situation of the glass capillary tube dipped in water and then broken at a certain height.

### Step 1: Understand the Initial Conditions
Initially, the water rises to a height of 9 cm in the capillary tube due to capillary action.

**Hint:** Recall that the height of the water column in a capillary tube is determined by the balance of adhesive and cohesive forces.

### Step 2: Break the Tube
The tube is then broken at a height of 5 cm. This means that the upper part of the tube (4 cm) is no longer in contact with the water.

**Hint:** Think about how breaking the tube affects the pressure and the height of the water column.

### Step 3: Determine the New Height of Water Column
After breaking the tube, the height of the water column will be equal to the height of the remaining part of the tube that is still submerged in water. Since the tube is broken at 5 cm, the height of the water column will now be 5 cm.

**Hint:** The height of the water column in the capillary tube is limited by the height of the tube that remains submerged.

### Step 4: Analyze the Angle of Contact
Initially, the angle of contact (θ1) was 0 degrees when the water rose to 9 cm. After breaking the tube, we need to find the new angle of contact (θ2).

Using the formula for capillary rise:
\[ h = \frac{2T \cos \theta}{r \rho g} \]
Where:
- \( h \) is the height of the water column,
- \( T \) is the surface tension,
- \( r \) is the radius of the tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step 5: Establish the Relationship Between Heights and Angles
From the capillary rise equation, we can establish the relationship:
\[ \frac{h_1}{\cos \theta_1} = \frac{h_2}{\cos \theta_2} \]
Where:
- \( h_1 = 9 \) cm (initial height),
- \( h_2 = 5 \) cm (new height),
- \( \theta_1 = 0 \) degrees (initial angle of contact).

Substituting the values:
\[ \frac{9}{\cos(0)} = \frac{5}{\cos \theta_2} \]
Since \( \cos(0) = 1 \):
\[ 9 = \frac{5}{\cos \theta_2} \]

### Step 6: Solve for \( \cos \theta_2 \)
Rearranging gives:
\[ \cos \theta_2 = \frac{5}{9} \]

### Step 7: Find \( \theta_2 \)
To find the angle of contact \( \theta_2 \):
\[ \theta_2 = \cos^{-1}\left(\frac{5}{9}\right) \]

### Final Results:
1. The height of the water column after breaking the tube is **5 cm**.
2. The angle of contact after breaking the tube is **\( \cos^{-1}\left(\frac{5}{9}\right) \)**.