A capillary tube is kept vertical with the lower end dipped in water . The height of water raised in the capillary is 4 cm . If the length of the capillary tube is made 2 cm , then the angle made by the water surface in the capillary with the wall is

Options

A. `0^@`
B. `30^@`
C. `60^@`
D. `90^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Capillary Rise
When a capillary tube is dipped in a liquid (in this case, water), the liquid rises in the tube due to surface tension. The height to which the liquid rises (h) is determined by the balance between the adhesive forces (between the liquid and the tube) and the cohesive forces (within the liquid).

### Step 2: Use the Capillary Rise Formula
The height of the liquid column in a capillary tube is given by the formula:

\[ h = \frac{2T \cos \theta}{r \rho g} \]

Where:
- \( h \) = height of the liquid column
- \( T \) = surface tension of the liquid
- \( \theta \) = angle of contact
- \( r \) = radius of the capillary tube
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity

### Step 3: Analyze the Given Data
From the problem:
- Initially, the height of water raised in the capillary tube is \( h_1 = 4 \, \text{cm} \).
- The new length of the capillary tube is \( h_2 = 2 \, \text{cm} \).

### Step 4: Establish the Relationship Between Heights and Angles
Since the surface tension and other parameters remain constant, we can establish the relationship between the heights and the cosines of the angles of contact:

\[ \frac{h_1}{h_2} = \frac{\cos \theta_1}{\cos \theta_2} \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:

\[ \frac{4}{2} = \frac{\cos \theta_1}{\cos \theta_2} \]

This simplifies to:

\[ 2 = \frac{\cos \theta_1}{\cos \theta_2} \]

### Step 6: Determine the Angle of Contact
From the previous analysis, we know that when the height was 4 cm, the angle of contact \( \theta_1 \) was such that \( \cos \theta_1 = 1 \) (which corresponds to a vertical rise). Therefore:

\[ \cos \theta_1 = 1 \implies \theta_1 = 0^\circ \]

Now substituting this into the equation:

\[ 2 = \frac{1}{\cos \theta_2} \]

This leads to:

\[ \cos \theta_2 = \frac{1}{2} \]

### Step 7: Calculate the Angle \( \theta_2 \)
Taking the inverse cosine:

\[ \theta_2 = \cos^{-1}\left(\frac{1}{2}\right) \]

This gives:

\[ \theta_2 = 60^\circ \]

### Final Answer
The angle made by the water surface in the capillary with the wall is \( 60^\circ \).

---