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 air above 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

The pressure exerted by the air above the mercury drop is the atmospheric pressure, P₀ = 1.0 10^5 Pa . The given surface area on the drop is A = 1 mm ^2 = 1.0 10⁻⁶ m ^2 . The force exerted by the air on this area is calculated as: F = P₀ A F = (1.0 10^5) (1.0 10⁻⁶) = 0.1 N The surface tension and radius of the drop are not required to find the force exerted by the outside air, as they are used to determine the pressure inside the drop.