Capillary Rise
Problems from IIT JEE
Problem (IIT JEE 2014):
A glass capillary tube is of the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $h$, where the radius of its cross-section is $b$. If the surface tension of water is $S$, its density is $\rho$, and its contact angle with glass is $\theta$, the value of $h$ will be (where $g$ is acceleration due to gravity),
- $\frac{2S}{b\rho g}\cos(\theta-\alpha)$
- $\frac{2S}{b\rho g}\cos(\theta+\alpha)$
- $\frac{2S}{b\rho g}\cos(\theta-\alpha/2)$
- $\frac{2S}{b\rho g}\cos(\theta+\alpha/2)$
Solution:
Let $R$ be the radius of the meniscus formed with a contact angle $\theta$ (see figure). By geometry, this radius makes an angle $\theta+\frac{\alpha}{2}$ with the horizontal and,
\begin{align}
\label{oxb:eqn:1}
\cos\left(\theta+\tfrac{\alpha}{2}\right)={b}/{R}.
\end{align}
Let $P_0$ be the atmospheric pressure and $P_1$ be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius $R$ is,
\begin{align}
\label{oxb:eqn:2}
P_0-P_1={2S}/{R}.
\end{align}
The hydrostatic pressure gives,
\begin{align}
\label{oxb:eqn:3}
P_0-P_1=h\rho g.
\end{align}
Eliminate $(P_0-P)$ from second and third equations and substitute $R$ from first equation to get,
\begin{align}
h=\frac{2S}{\rho g R}=\frac{2S}{b\rho g}\cos\left(\theta+\tfrac{\alpha}{2}\right).\nonumber
\end{align}