Question

Figure shows a capillary tube of radius $r$ dipped into water. If the atmospheric pressure is $P_0$, the pressure at point $A$ ( just below the meniscus ) is

• A. $P_0$
• B. $P_0+\frac{2S}{r}$
• C. $P_0-\frac{2S}{r}$
• D. $P_0-\frac{4S}{r}$

Answer 1

Answer: (C)

$P_0-\frac{2S}{r}$

The pressure just above the meniscus is atmospheric pressure and is greater than the pressure just below the meniscus since the pressure on the concave side is greater than that on the convex side.
$P_{concave}-P_{convex}=S(\frac{1}{R_1}+\frac{1}{R_2})$
For a spherical surface $R_1=R_2=r$
Hence, $P_{concave}-P_{convex}=\frac{2S}{r}$
In the given scenario, $P_{concave}=P_0$ where $P_0$ is the atmospheric pressure.
$P_{convex}=P_0-\frac{2S}{r}$