Question

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 = 1.0 × 10⁵ Pa and surface tension of mercury = 0.465 N m⁻¹. Neglect the effect of gravity. Assume all numbers to be exact.

Answer 1

Given:
Surface area of mercury drop, A = 1 mm² = 10⁻⁶ m²
Radius of mercury drop, r = 4 mm = 4 × 10⁻³ m
Atmospheric pressure, P₀ = 1.0 × 10⁵ P_a
Surface tension of mercury, T = 0.465 N/m

(a) Force exerted by air on the surface area:
F = P₀A
⇒ F= 1.0 × 10⁵ × 10⁻⁶ = 0.1 N

(b) Force exerted by mercury below the surface area:
$$\begin{gathered} \mathrm{Pressure}P\text{'}=P_0+\frac{2T}{r} \\ F=P\text{'}A=\left(P_0+\frac{2T}{r}\right)A \\ =\left(0.1+\frac{2\times 0.465}{4\times {10}^{-3}}\right)\times {10}^{-6} \\ =0.1+0.00023=0.10023\mathrm{N} \end{gathered}$$

(c) Force exerted by mercury surface in contact with it:
$$\begin{gathered} P=\frac{2T}{r} \\ F=PA=\frac{2T}{r}A \\ =\frac{2\times 0.465}{4\times {10}^{-3}}\times {10}^{-6}=0.00023\mathrm{N} \end{gathered}$$