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 below 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 inside the mercury drop is given by P = P₀ + 2T R , where P₀ is the atmospheric pressure, T is the surface tension, and R is the radius of the drop. Given: P₀ = 1.0 10^5 Pa T = 0.465 N/m R = 4.0 mm = 4.0 10⁻³ m The excess pressure inside the drop is: P = 2T R = 2 0.465 4.0 10⁻³ = 232.5 Pa The total pressure of the mercury just inside the surface is: P = P₀ + P = 100000 + 232.5 = 100232.5 Pa The force exerted by the mercury below the small surface area A = 1 mm ^2 = 1.0 10⁻⁶ m ^2 is due to this internal