Solution:

Given, each side of regular polygon of n sides has a resistance R.

Equivalent resistance between any two adjacent vertices is to be found.

Let's consider a polygon of n sides with vertices A1, A2, A3, ..., An.

To find the equivalent resistance between any two adjacent vertices, let's take two adjacent vertices A1 and A2.

We will convert this network into a simpler network with equivalent resistance.

We will replace the sides connecting A1 and A2 by a single equivalent resistance, say R1.

Now, let's consider another pair of adjacent vertices A2 and A3.

We can replace the sides connecting A2 and A3 by a single equivalent resistance, say R2.

Similarly, we can replace the sides connecting A3 and A4 by a single equivalent resistance, say R3.

We can continue this process till we replace the sides connecting An and A1 by a single equivalent resistance, say Rn.

Now, we have a simpler network with n equivalent resistances.

To find the equivalent resistance between A1 and A2, we need to find the equivalent resistance of the network consisting of R1, R2, R3, ..., Rn.

Let's calculate the equivalent resistance of this network.

The equivalent resistance of resistors in series is the sum of their individual resistances.

Hence, the equivalent resistance of R1, R2, R3, ..., Rn is given by:

R_eq = R1 + R2 + R3 + ... + Rn

We know that the resistance of each side is R.

Hence, R1 = R2 = R3 = ... = Rn = R.

Substituting in the above equation, we get:

R_eq = R + R + R + ... + R (n times)

R_eq = nR

Therefore, the equivalent resistance between any two adjacent vertices is nR.

Option (C) states that the minimum value of equivalent resistance is R(n-1)/n^2.

Let's check if this is true.

We can write nR as R(1+1+1+...+1) (n times)

Hence, we need to find the minimum value of (1+1+1+...+1) (n times) / n^2.

The numerator is simply n.

Hence, we need to find the minimum value of n/n^2 = 1/n.

Therefore, the minimum value of equivalent resistance is R/n.

Hence, option (C) is the correct answer.

C