Question

A water drop of $0.01\text{ cm}^3$ is squeezed between two glass plates and spreads in to area of $10\text{ cm}^2$ . If surface tension of water is $70\text{ dyne /cm}$ then the normal force required to separate glass plates from each other will be}

• A. 12 N
• B. 14 N
• C. 16 N
• D. 28 N

Hint:

For a liquid film between plates, remember $F = \frac{2A^2T}{V}$. This formula accounts for both surfaces of the film.

Answer 1

The Correct Option is B

Step 1: Concept
The force required to separate the plates is $F = \frac{2AT}{d}$, where $A$ is area, $T$ is surface tension, and $d$ is the thickness of the film.

Step 2: Meaning
Volume $V = A \times d \implies d = V/A$. Substituting $d$ into the force formula gives $F = \frac{2A^2T}{V}$.

Step 3: Analysis
Given: $A = 10 \text{ cm}^2$, $V = 0.01 \text{ cm}^3$, $T = 70 \text{ dyne/cm}$. $F = \frac{2 \times (10)^2 \times 70}{0.01} = \frac{2 \times 100 \times 70}{0.01} = 1400000 \text{ dynes}$. Convert to Newtons: $1 \text{ N} = 10^5 \text{ dynes}$. $F = \frac{14 \times 10^5}{10^5} = 14 \text{ N}$.

Step 4: Conclusion
The normal force required is $14 \text{ N}$. Final Answer: (B)