Two glass plates are separated by water. If surface tension of water is `75 dyn//cm` and the area of each plate wetted by water is `8 cm^(2)` and the distance between the plates is `0.12 mm`, then the force applied to separate the two plates is

Options

A. `10^(2)dyn`
B. `10^(4)dyn`
C. `10^(5)dyn`
D. `10^(6)dyn`

Explanation

To find the force applied to separate the two glass plates that are separated by water, we can use the concept of surface tension. The force due to surface tension can be calculated using the formula:

\[ F = 2 \times \gamma \times L \]

Where:
- \( F \) is the force,
- \( \gamma \) is the surface tension of the liquid (water in this case),
- \( L \) is the length of the wetted perimeter in contact with the liquid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Surface tension of water, \( \gamma = 75 \, \text{dyn/cm} \)
- Area of each plate, \( A = 8 \, \text{cm}^2 \)
- Distance between the plates, \( d = 0.12 \, \text{mm} = 0.012 \, \text{cm} \) (conversion from mm to cm)

2. **Calculate the wetted length \( L \):**
- Since the area \( A \) is given, we can find the wetted length \( L \) using the formula:
\[ A = L \times d \]
Rearranging gives:
\[ L = \frac{A}{d} \]
Substituting the values:
\[ L = \frac{8 \, \text{cm}^2}{0.012 \, \text{cm}} = \frac{8}{0.012} \approx 666.67 \, \text{cm} \]

3. **Calculate the force \( F \):**
- Now we can substitute \( L \) and \( \gamma \) into the force formula:
\[ F = 2 \times \gamma \times L = 2 \times 75 \, \text{dyn/cm} \times 666.67 \, \text{cm} \]
\[ F = 150 \times 666.67 \approx 100000 \, \text{dyn} \]

4. **Convert the force to appropriate units if necessary:**
- The force calculated is already in dynes, which is suitable for this context.

5. **Final Answer:**
- The force applied to separate the two plates is approximately \( 100000 \, \text{dyn} \).