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 ($g$ is the acceleration due to gravity)

Step 1: Identify the geometry of the capillary tube.

The capillary tube is a truncated cone with apex angle $\alpha$, meaning the walls are tilted from the vertical by the semi-angle.

$$\text{Semi-angle of the cone} = \frac{\alpha}{2}$$

Step 2: Determine the angle of the meniscus tangent with the vertical.

The contact angle of water with the glass wall is $\theta$, and since the wall is tilted by $\frac{\alpha}{2}$ from the vertical, the tangent to the liquid meniscus at the contact line makes an angle of $\theta + \frac{\alpha}{2}$ with the vertical.

$$\phi = \theta + \frac{\alpha}{2}$$

Step 3: Relate the radius of curvature of the meniscus to the tube radius.

Let $R$ be the radius of curvature of the spherical meniscus, and $b$ be the radius of the capillary tube at height $h$.

$$b = R \cos\left(\theta + \frac{\alpha}{2}\right)$$ $$R = \frac{b}{\cos\left(\theta + \frac{\alpha}{2}\right)}$$

Step 4: Apply the pressure balance equation.

The pressure difference across the curved meniscus is balanced by the hydrostatic pressure of the water column of height $h$.

$$\Delta P = \frac{2S}{R}$$ $$\rho g h = \frac{2S}{R}$$

Step 5: Solve for the height $h$.

Substitute the expression for $R$ into the pressure balance equation.

$$\rho g h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b}$$ $$h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b \rho g}$$

Answer:

$$h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b \rho g}$$