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`        Consider a small surface area of 1 mm2 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 tension of mercury it. Atmospheric pressure = 1.0 × 105 Pa and surface tension of mercury = 0.465 N m-1. Neglect the effect of gravity. Assume all numbers to be exact. `
3 years ago

Kevin Nash
332 Points
```										Sol. a) F = P0A
b) Pressure = p0 + (2T/r)
F = P’A = (P0 + (2T/r)A
c) P = 2T/r
F = PA = 2T/r A

```
3 years ago
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