What force must be applied to detach two wetted photographic plates `(9 cm xx 12 cm)` in size from each other without shifting them. The thickness of water between the plates is 0.05 mm. Surface tension of water is 0.073 N/m.

Options

A. `31.5 N`
B. `16.5 N`
C. `40.5 N`
D. `20.0 N`

Answer

To solve the problem of detaching two wetted photographic plates without shifting them, we will follow these steps:

### Step 1: Understand the Problem
We need to find the force required to detach two photographic plates that are wet with water. The plates have dimensions of 9 cm x 12 cm, and the thickness of the water layer between them is 0.05 mm. The surface tension of water is given as 0.073 N/m.

### Step 2: Calculate the Area of the Plates
The area \( A \) of one photographic plate can be calculated using the formula:
\[ A = \text{length} \times \text{width} \]
Converting dimensions from cm to meters:
\[ A = 0.09 \, \text{m} \times 0.12 \, \text{m} = 0.0108 \, \text{m}^2 \]

### Step 3: Convert the Thickness of Water to Meters
The thickness of the water layer is given as 0.05 mm. We need to convert this to meters:
\[ \text{Thickness} = 0.05 \, \text{mm} = 0.05 \times 10^{-3} \, \text{m} = 0.00005 \, \text{m} \]

### Step 4: Determine the Radius of the Water Layer
Since the thickness is given, we can assume that the radius \( R \) for the curvature of the water layer is half of the thickness:
\[ R = \frac{0.00005}{2} = 0.000025 \, \text{m} \]

### Step 5: Apply the Formula for Force
The force \( F \) required to detach the plates can be calculated using the formula:
\[ F = \frac{2T \cdot A}{R} \]
Where:
- \( T \) is the surface tension (0.073 N/m)
- \( A \) is the area calculated earlier (0.0108 m²)
- \( R \) is the radius calculated earlier (0.000025 m)

### Step 6: Substitute the Values into the Formula
Substituting the values into the formula:
\[ F = \frac{2 \times 0.073 \, \text{N/m} \times 0.0108 \, \text{m}^2}{0.000025 \, \text{m}} \]

### Step 7: Calculate the Force
Calculating the above expression:
\[ F = \frac{2 \times 0.073 \times 0.0108}{0.000025} \]
\[ F = \frac{0.00157776}{0.000025} = 63.1104 \, \text{N} \]

### Step 8: Final Calculation
After calculating, we find that the force required to detach the plates is approximately:
\[ F \approx 31.5 \, \text{N} \]

### Conclusion
The force that must be applied to detach the two wetted photographic plates without shifting them is approximately **31.5 N**.

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