Question

Figure shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P₀, the pressure at point A is
(a) P₀
(b) $P_0+\frac{2S}{r}$
(c) $P_0-\frac{2S}{r}$
(d) $P_0-\frac{4S}{r}$

Answer 1

Correct option: (c)



$$\begin{gathered} \mathrm{Here}: \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{tube}=\mathrm{r} \\ \mathrm{Net}\mathrm{upward}\mathrm{force}\mathrm{due}\mathrm{to}\mathrm{surface}\mathrm{tension}=\mathrm{Scos\theta }\times 2\mathrm{\pi r} \\ \mathrm{Upward}\mathrm{pressure}=\frac{\mathrm{Scos\theta }\times 2\mathrm{\pi r}}{{\mathrm{\pi r}}^2}=\frac{2\mathrm{Scos\theta }}{\mathrm{r}} \\ \mathrm{Net}\mathrm{downward}\mathrm{pressure}\mathrm{due}\mathrm{to}\mathrm{atmosphere}={\mathrm{P}}_{\mathrm{o}} \\ \Rightarrow \mathrm{Net}\mathrm{pressure}\mathrm{at}\mathrm{A}={\mathrm{P}}_{\mathrm{o}}-\frac{2\mathrm{Scos\theta }}{\mathrm{r}} \\ \mathrm{Since}\mathrm{\theta }\mathrm{is}\mathrm{small}, \\ \mathrm{cos\theta }\approx 1. \\ \Rightarrow \mathrm{Net}\mathrm{pressure}={\mathrm{P}}_{\mathrm{o}}-\frac{2\mathrm{S}}{\mathrm{r}} \end{gathered}$$