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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.
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Solution
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 Pa
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 :
\[\text{ Pressure P' = P}_0 + \frac{2T}{\text{r}}\]
\[F = P'A = \left( P_0 + \frac{2T}{r} \right)A\]
\[ = \left( 0 . 1 + \frac{2 \times 0 . 465}{4 \times {10}^{- 3}} \right) \times {10}^{- 6} \]
\[ = 0 . 1 + 0 . 00023 = 0 . 10023 \text{N}\]
(c) Force exerted by mercury surface in contact with it:
\[P = \frac{2T}{r}\]
\[F = PA = \frac{2T}{r}A\]
\[ = \frac{2 \times 0 . 465}{4 \times {10}^{- 3}} \times {10}^{- 6} = 0 . 00023 \text{ N}\]
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