A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

The correct option/answer is: R/16

Let's break down the problem step-by-step to calculate the net effective resistance of the combination:

Step 1: Determine the resistance of each cut piece
Initially, we have a wire of total resistance $R$. The resistance of a uniform wire is directly proportional to its length. When the wire is cut into 8 equal pieces, the resistance of each individual piece, let's call it $r_{\mathrm{piece}}$, is:
$r_{\mathrm{piece}}=\frac{R}{8}$

Step 2: Find the equivalent resistance of each parallel combination
We are told that two separate sets are made. Each set is formed by connecting four of these pieces in parallel.
Let $R_p$ be the equivalent resistance of one such set consisting of 4 pieces connected in parallel.
The formula for the equivalent resistance of $n$ identical resistors in parallel is:
$\frac{1}{R_p}=\frac{1}{r_{\mathrm{piece}}}+\frac{1}{r_{\mathrm{piece}}}+\frac{1}{r_{\mathrm{piece}}}+\frac{1}{r_{\mathrm{piece}}}=\frac{4}{r_{\mathrm{piece}}}$
Taking the reciprocal, we get:
$R_p=\frac{r_{\mathrm{piece}}}{4}$
Substituting $r_{\mathrm{piece}}=\frac{R}{8}$ into this equation:
$R_p=\frac{R/8}{4}=\frac{R}{32}$

Step 3: Calculate the net effective resistance of the series combination
Now, these two sets (each having an equivalent resistance of $R_p$) are connected in series with each other.
For a series combination, the net effective resistance $R_{\mathrm{net}}$ is the sum of the individual resistances:
$R_{\mathrm{net}}=R_p+R_p=2R_p$
Substituting the value of $R_p$ we calculated in Step 2:
$R_{\mathrm{net}}=2\cdot \left(\frac{R}{32}\right)=\frac{R}{16}$

Therefore, the net effective resistance of the final combination is R/16.