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
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 it. Atmospheric pressure = 1.0x105 Pa and surface tension of mercury = 0.465 N/m. Neglect the effect of gravity. Assume all numbers to be exact.
1 Answers
The area of the given surface, A = 1 mm² =1x10⁻⁶ m².
(a) The force exerted by the air above it =A*atmospheric pressure
=1x10⁻⁶ m²*1.0x10⁵ N/m²
=0.10 N
(b) The surface tension of the mercury, S =0.465 N/m
The excess pressure inside the drop, Δp = 2S/r
→Δp =2*0.465/(4/1000) =465/2 = 232.5 N/m²
Total pressure inside the drop =Δp + Atmospheric pressure
=232.5 +1.0x10⁵ N/m²
Force on the area by the mercury =A*mercury pressure
=1x10⁻⁶(232.5 +1.0x10⁵) N/m²
=2.325x10⁻⁴+0.10 N/m²
=0.00023 + 0.10 N/m²
=0.10023 N
(c) The mercury surface in contact with the area has a net pressure equal to the pressure in mercury minus the atmospheric pressure outside, i.e. excess pressure inside the drop =2S/r
=2*0.465/(4/1000) N/m²
=465/2 N/m²
=232.5 N/m²
Hence the force = Area*Pressure
=1x10⁻⁶*232.5 N
=0.00023 N