Key idea: capillary rise is found by balancing forces

In a capillary tube, the upward vertical component of surface tension balances the weight of the liquid column.

Geometry of the tube

The capillary is a truncated cone with apex angle $\alpha$.

Hence, the wall of the cone makes an angle $\frac{\alpha}{2}$ with the axis of the tube.

At the level where the liquid rises, the radius of cross-section is given as $b$.

Surface tension force

Surface tension acts tangentially to the liquid surface at the line of contact.

For an ordinary cylindrical tube, the vertical component is $2\pi r S \cos\theta$ because the wall is parallel to the axis.

Here, the wall is inclined. Since the wall makes angle $\alpha/2$ with the vertical axis, the effective angle between the surface tension direction and the vertical changes.

Therefore, the vertical component becomes $2\pi b S \cos\left(\theta + \frac{\alpha}{2}\right).$

Weight of the liquid column

If the liquid rises to height $h$, then the weight of the liquid column is approximately $\pi b^2 h\rho g.$

Force balance

At equilibrium, $2\pi b S \cos\left(\theta + \frac{\alpha}{2}\right)=\pi b^2 h\rho g.$

Cancel $\pi b$ from both sides: $2S\cos\left(\theta + \frac{\alpha}{2}\right)=b h\rho g.$

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

Matching with options

This corresponds to: $\boxed{\frac{2S}{b\rho g}\cos\left(\theta + \frac{\alpha}{2}\right)}$

So the correct option is D.