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: Understanding the Concept:
When a liquid drop is squeezed between two flat plates, it forms a thin film. The surface tension of the liquid creates a pressure difference across the curved free surface of the film (meniscus) at the edges, leading to an attractive force between the plates.
Step 2: Key Formula or Approach:
The normal force $F$ required to separate the two plates is given by:
\[ F = \frac{2TA}{t} \]
where $T$ is the surface tension, $A$ is the wetted area, and $t$ is the thickness of the film.
Since the volume $V$ of the drop remains constant, $V = A \times t$, which means $t = \frac{V}{A}$.
Substituting this into the force equation:
\[ F = \frac{2TA}{\frac{V}{A}} = \frac{2TA^2}{V} \]
Step 3: Detailed Explanation:
Given values (let's use CGS units first for calculation, then convert to SI):
Volume, $V = 0.01\text{ cm}^3 = 10^{-2}\text{ cm}^3$
Area, $A = 10\text{ cm}^2$
Surface tension, $T = 70\text{ dyne/cm}$
Substitute these into the derived formula:
\[ F = \frac{2 \times 70 \times (10)^2}{10^{-2}} \]
\[ F = \frac{140 \times 100}{10^{-2}} \]
\[ F = 14000 \times 10^2 \]
\[ F = 1,400,000\text{ dynes} = 14 \times 10^5\text{ dynes} \]
To convert dynes to Newtons (SI unit), we use the relation $1\text{ N} = 10^5\text{ dynes}$.
\[ F = \frac{14 \times 10^5}{10^5}\text{ N} \]
\[ F = 14\text{ N} \]
Step 4: Final Answer:
The normal force required is $14\text{ N}$.