Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity $$\rho$$. The resistance between the two spheres will be:

We are given two concentric conducting spheres. Their radii are $$a$$ for the inner sphere and $$b$$ for the outer sphere, where, as stated, $$b > a$$. The space between them is completely filled with a material of resistivity $$\rho$$. Our task is to calculate the electrical resistance offered by this spherical shell of material when current flows radially from one sphere to the other.

To find resistance, we first recall the basic definition that connects resistivity with resistance for a small (differential) element:

$$dR=\rho\,\dfrac{dl}{A}$$

Here, $$dR$$ is the infinitesimal resistance of a small slab of material, $$dl$$ is its length along the direction of current, and $$A$$ is the cross-sectional area perpendicular to that current. We will now apply this formula to a thin spherical shell lying between the radii $$r$$ and $$r+dr$$.

For a radial current in a sphere, the current lines are directed along the radius, so:

• The infinitesimal length is simply $$dl = dr$$.
• The cross-sectional area through which current passes at radius $$r$$ is the entire spherical surface at that radius, whose area equals $$4\pi r^2$$.

Substituting these into the formula, we obtain the elemental resistance:

$$dR=\rho\,\dfrac{dr}{4\pi r^2}$$

Now we must sum (integrate) all these differential resistances from the inner radius $$a$$ up to the outer radius $$b$$, because each thin shell contributes in series along the radial path:

$$R=\displaystyle\int_a^{b}dR=\int_a^{b}\rho\,\dfrac{dr}{4\pi r^2}$$

The constants $$\rho$$ and $$4\pi$$ can be pulled outside the integral:

$$R=\dfrac{\rho}{4\pi}\,\int_a^{b}r^{-2}\,dr$$

We now integrate $$r^{-2}$$. The standard integral formula is:

$$\displaystyle\int r^{-2}\,dr = -\dfrac{1}{r}+C$$

Applying the limits from $$a$$ to $$b$$, we have:

$$\int_a^{b}r^{-2}\,dr = \left[-\dfrac{1}{r}\right]_{a}^{b} = \left(-\dfrac{1}{b}\right) - \left(-\dfrac{1}{a}\right) = \dfrac{1}{a}-\dfrac{1}{b}$$

Substituting this result back into our expression for $$R$$ yields:

$$R=\dfrac{\rho}{4\pi}\,\left(\dfrac{1}{a}-\dfrac{1}{b}\right)$$

This formula matches exactly with Option C in the given list.

Hence, the correct answer is Option C.