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The cylinder tapers from end to end. Since the radius increases linearly, we have the rate of increase of radius per unit length of the cylinder as dr = $\frac{b-a}{L}$

Now, consider a thin strip of thickness dx, at a distance x from the end with radius a. Its radius is a + xdr, or, its radius is $\frac{b-a}{L} x + a$

Resistance of this element is: dR = $\frac{\rho dx}{A}$ , where A is the elemental area. Hence, on substituting the value for A, we get:

dR = $\frac{\rho dx}{\pi {(a+\frac{b-a}{L} x})^2}$

The overall resistance is got by integrating this under proper limits. For sake of simplicity, set k = $\frac{b-a}{L}$

R = $\int^L_0 \frac{\rho}{\pi (a+kx)^2} \, dx$

On integrating and putting the limits, we get R = $\frac{\rho L}{\pi ab}$ units.

Oh yes, good work. I made a grave mistake there.

Here's how we go about it:

Considering an element of thickness dx, the charge on it = $\alpha xdx$ . Hence, the Potential due to this charge at x = -d is nothing but $V = \frac{K dq}{x+d}$ .

The potential due to the whole charge will simply be $\int^{L+d}_{d} \frac{K \alpha x}{x+d} \, dx$ where K = $\frac{1}{4 \pi \epsilon_o}$ .

A straight forward sum that makes use of the formula:

$h = \frac{2S cos \theta}{r \rho g}$

S = Surface Tension of the liquid, r = inner radius of the capillary tube, etc. So go ahead!