A drop of some liquid of volume `0.04 cm^(3)` is placed on the surface of a glass slide. Then, another glass forms a thin layer of area `20 cm^(2)` between the surfaces of the two slides. To separate the slides a force of `16 xx 10^(5)` dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne `cm^(-1))`

Options

A. 60
B. 70
C. 80
D. 90

Explanation

To solve the problem step by step, we will follow the logical sequence of calculations based on the information given.

### Step 1: Understand the Given Information
We have:
- Volume of the liquid drop, \( V = 0.04 \, \text{cm}^3 \)
- Area of the thin layer between the slides, \( A = 20 \, \text{cm}^2 \)
- Force required to separate the slides, \( F = 16 \times 10^5 \, \text{dyne} \)

### Step 2: Calculate the Length of the Liquid Layer
We know that the volume of the liquid is given by the formula:
\[ V = A \times h \]
where \( h \) is the height (or thickness) of the liquid layer. Rearranging the formula gives:
\[ h = \frac{V}{A} \]
Substituting the values:
\[ h = \frac{0.04 \, \text{cm}^3}{20 \, \text{cm}^2} = 0.002 \, \text{cm} \]

### Step 3: Calculate the Radius of Curvature
The radius of curvature \( R \) for a drop of liquid can be approximated as half the height of the liquid layer:
\[ R = \frac{h}{2} = \frac{0.002 \, \text{cm}}{2} = 0.001 \, \text{cm} \]

### Step 4: Calculate the Excess Pressure
The excess pressure \( P \) across the liquid layer due to surface tension is given by:
\[ P = \frac{F}{A} \]
Substituting the values:
\[ P = \frac{16 \times 10^5 \, \text{dyne}}{20 \, \text{cm}^2} = 8 \times 10^4 \, \text{dyne/cm}^2 \]

### Step 5: Relate Excess Pressure to Surface Tension
The relationship between excess pressure and surface tension \( S \) is given by:
\[ P = \frac{S}{R} \]
Rearranging this gives:
\[ S = P \times R \]

### Step 6: Substitute the Values to Find Surface Tension
Now substituting the values of \( P \) and \( R \):
\[ S = (8 \times 10^4 \, \text{dyne/cm}^2) \times (0.001 \, \text{cm}) = 80 \, \text{dyne/cm} \]

### Conclusion
Thus, the surface tension of the liquid is:
\[ \boxed{80 \, \text{dyne/cm}} \]