A drop of liquid pressed between two glass plates spreads to a circle of diameter 10 cm. Thickness of the liquid film is 0.5 mm and surface tension is `70xx10^(-3)Nm^(-1)` the force required to pull them apart is

Options

A. 4.4N
B. 1.1N
C. 2.2N
D. 3.6N

Explanation

To solve the problem, we need to calculate the force required to pull apart the two glass plates that are pressed together by a liquid film. We'll use the formula for the force due to surface tension.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Diameter of the liquid film (D) = 10 cm = 0.1 m
- Thickness of the liquid film (t) = 0.5 mm = 0.5 × 10^(-3) m
- Surface tension (T) = 70 × 10^(-3) N/m

2. **Calculate the Radius of the Circle**:
- The radius (r) is half of the diameter.
\[ r = \frac{D}{2} = \frac{0.1}{2} = 0.05 \, \text{m} \]

3. **Calculate the Area of the Circle**:
- The area (A) of the circle is given by the formula \( A = \pi r^2 \).
\[ A = \pi (0.05)^2 = \pi \times 0.0025 \approx 0.007854 \, \text{m}^2 \]

4. **Calculate the Force Due to Surface Tension**:
- The force (F) due to surface tension acting on both surfaces of the liquid film can be calculated using the formula:
\[ F = 2 \times T \times \frac{A}{t} \]
- Substituting the values:
\[ F = 2 \times (70 \times 10^{-3}) \times \frac{0.007854}{0.5 \times 10^{-3}} \]

5. **Simplify the Calculation**:
- First, calculate \( \frac{A}{t} \):
\[ \frac{0.007854}{0.5 \times 10^{-3}} = \frac{0.007854}{0.0005} = 15.708 \]
- Now substitute this back into the force equation:
\[ F = 2 \times (70 \times 10^{-3}) \times 15.708 \]
- Calculate the force:
\[ F = 2 \times 70 \times 15.708 \times 10^{-3} = 2200.88 \times 10^{-3} \, \text{N} \approx 2.2 \, \text{N} \]

6. **Final Result**:
- The force required to pull the glass plates apart is approximately **2.2 N**.