Two capillary tubes P and Q are dipped in water. The height of water level in capillary P is 2/3 to the height in Q capillary. The ratio of their diameters is

Options

A. `2 : 3`
B. ` 3 :2`
C. `3:4`
D. `4:3`

Explanation

To solve the problem, we need to understand the relationship between the height of water in capillary tubes and their diameters. The height of liquid 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-by-Step Solution:

1. **Identify the relationship between height and radius**:
From the formula, we can see that the height \( h \) is inversely proportional to the radius \( r \) of the capillary tube. This means that if the height of the liquid rises in one tube, the radius must be smaller, and vice versa.

2. **Set up the ratio of heights**:
Let \( h_P \) be the height of water in capillary tube P and \( h_Q \) be the height of water in capillary tube Q. According to the problem, we have:
\[ h_P = \frac{2}{3} h_Q \]

3. **Express the radii in terms of heights**:
Since the radius is inversely proportional to the height, we can express the ratio of the radii of the two capillaries as:
\[ \frac{r_P}{r_Q} = \frac{h_Q}{h_P} \]

4. **Substitute the height relationship**:
Substitute \( h_P \) into the ratio:
\[ \frac{r_P}{r_Q} = \frac{h_Q}{\frac{2}{3} h_Q} = \frac{h_Q \cdot 3}{2 h_Q} = \frac{3}{2} \]

5. **Relate the radius to the diameter**:
The diameter \( d \) is twice the radius \( r \):
\[ d_P = 2r_P \quad \text{and} \quad d_Q = 2r_Q \]
Therefore, the ratio of the diameters is the same as the ratio of the radii:
\[ \frac{d_P}{d_Q} = \frac{r_P}{r_Q} = \frac{3}{2} \]

6. **Final answer**:
The ratio of the diameters of the capillary tubes P and Q is:
\[ \frac{d_P}{d_Q} = \frac{3}{2} \]

### Conclusion:
The ratio of the diameters of capillary tubes P and Q is \( \frac{3}{2} \).