Consider a small surface area of 1 mm ^2 at the top of a mercury drop of radius 4.0 mm . Determine the force exerted on this area by the mercury surface in contact with it. Atmospheric pressure is 1.0 10^5 Pa and surface tension of mercury is 0.465 N/m . Neglect the effect of gravity. Assume all numbers to be exact.

Step-by-step solution

For a small area A on the surface of a liquid drop in equilibrium, the outward force due to the pressure inside the drop is balanced by the inward force due to atmospheric pressure and the inward resultant force exerted by the surrounding surface due to surface tension. Let P_ in be the pressure inside the drop and P₀ be the atmospheric pressure. The excess pressure is given by: P_ ex = P_ in - P₀ = 2T R The force exerted by the surrounding mercury surface on the small area A is the resultant of the surface tension