A thin square plate of side `5 cm` is suspended vertically a balance so that lower side just dips into water with side to surface. When the plate is clean `(theta=0^@)`, it appears to weigh `0.044 N`. But when the plate is greasy `(theta=180^@)` it appears to weigh `0.03 N`. The surface tension of water is

Options

A. `3.5xx10^(-2)N//m`
B. `7.0xx10^(-2)N//m`
C. `14.0xx10^(-2)N//m`
D. `1.08N//m`

Explanation

To find the surface tension of water based on the given problem, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**:
- Side of the square plate, \( a = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Weight of the plate when clean, \( W = 0.044 \, \text{N} \)
- Apparent weight of the plate when greasy, \( W' = 0.03 \, \text{N} \)

2. **Calculate the perimeter of the plate**:
The perimeter \( P \) of the square plate is given by:
\[ P = 4a = 4 \times 0.05 \, \text{m} = 0.2 \, \text{m} \]

3. **Understand the forces acting on the plate**:
- When the plate is clean, the weight \( W \) is balanced by the spring force (the reading on the balance).
- When the plate is greasy, the apparent weight \( W' \) is reduced due to the upward force from the surface tension.

4. **Set up the equation for the greasy plate**:
The surface tension force \( F_s \) acts upwards and can be expressed as:
\[ F_s = T \times P \]
where \( T \) is the surface tension and \( P \) is the perimeter of the plate.

The relationship when the plate is greasy can be written as:
\[ W' = W - F_s \]
Substituting for \( F_s \):
\[ W' = W - T \times P \]

5. **Rearranging the equation**:
Rearranging the above equation gives:
\[ T \times P = W - W' \]
Therefore:
\[ T = \frac{W - W'}{P} \]

6. **Substituting the values**:
Now we can substitute the known values into the equation:
\[ T = \frac{0.044 \, \text{N} - 0.03 \, \text{N}}{0.2 \, \text{m}} = \frac{0.014 \, \text{N}}{0.2 \, \text{m}} = 0.07 \, \text{N/m} \]

7. **Final answer**:
The surface tension of water is:
\[ T = 0.07 \, \text{N/m} \]