Consider a small surface area of `1 mm^2` at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure `=1.0xx10^5Pa` and surface tension of mercury `=0.465Nm^-1.` Neglect the effect of gravity. Assume all numbers to be exact.

Explanation

To solve the problem, we will calculate the forces exerted on a small surface area of `1 mm²` at the top of a mercury drop of radius `4.0 mm`. We will find the forces exerted by the air above, the mercury below, and the mercury surface in contact with it.

### Given Data:
- Surface area, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Radius of mercury drop, \( R = 4.0 \, \text{mm} = 4.0 \times 10^{-3} \, \text{m} \)
- Atmospheric pressure, \( P_0 = 1.0 \times 10^5 \, \text{Pa} \)
- Surface tension of mercury, \( \gamma = 0.465 \, \text{N/m} \)

### (a) Force exerted by the air above the mercury drop:
The force exerted by the air can be calculated using the formula:
\[ F_{\text{air}} = P_0 \times A \]
Substituting the values:
\[ F_{\text{air}} = (1.0 \times 10^5 \, \text{Pa}) \times (1 \times 10^{-6} \, \text{m}^2) = 0.1 \, \text{N} \]

### (b) Force exerted by the mercury below the drop:
To find the force exerted by the mercury below, we first need to calculate the excess pressure inside the mercury drop using the formula:
\[ \Delta P = \frac{2\gamma}{R} \]
Substituting the values:
\[ \Delta P = \frac{2 \times 0.465 \, \text{N/m}}{4.0 \times 10^{-3} \, \text{m}} = 232.5 \, \text{Pa} \]
The total pressure below the surface is the sum of the atmospheric pressure and the excess pressure:
\[ P_{\text{total}} = P_0 + \Delta P = (1.0 \times 10^5 \, \text{Pa}) + (232.5 \, \text{Pa}) = 100232.5 \, \text{Pa} \]
Now, we can calculate the force exerted by the mercury below:
\[ F_{\text{mercury}} = P_{\text{total}} \times A \]
Substituting the values:
\[ F_{\text{mercury}} = (100232.5 \, \text{Pa}) \times (1 \times 10^{-6} \, \text{m}^2) = 0.1002325 \, \text{N} \]

### (c) Force exerted by the mercury surface in contact with the area:
The force exerted by the mercury surface in contact can be calculated using the excess pressure:
\[ F_{\text{contact}} = \Delta P \times A \]
Substituting the values:
\[ F_{\text{contact}} = (232.5 \, \text{Pa}) \times (1 \times 10^{-6} \, \text{m}^2) = 0.0002325 \, \text{N} \]

### Summary of Forces:
- Force exerted by air above: \( F_{\text{air}} = 0.1 \, \text{N} \)
- Force exerted by mercury below: \( F_{\text{mercury}} = 0.1002325 \, \text{N} \)
- Force exerted by mercury surface in contact: \( F_{\text{contact}} = 0.0002325 \, \text{N} \)