A wire of length ‘l’ and resistance 100Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

The correct option/answer is 52Ω.

Let us break down the solution step-by-step:

Step 1: Find the resistance of each part.
The original wire has a length $l$ and a total resistance $R=100\Omega$. The wire is divided into 10 equal parts. Since the resistance of a uniform wire is directly proportional to its length, the resistance of each of the 10 equal parts is:

$R_{\text{part}}=\frac{100}{10}=10\Omega$

Step 2: Calculate the resistance of the first 5 parts in series.
The first 5 parts (each of resistance $10\Omega$) are connected in series. The equivalent resistance $R_s$ for this series combination is:

$R_s=10+10+10+10+10=5\times 10=50\Omega$

Step 3: Calculate the resistance of the next 5 parts in parallel.
The remaining 5 parts (each of resistance $10\Omega$) are connected in parallel. The equivalent resistance $R_p$ for this parallel combination is:

$\frac{1}{R_p}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{5}{10}=\frac{1}{2}$

Hence, taking the reciprocal:
$R_p=2\Omega$

Step 4: Find the resistance of the final combination.
These two groups (the series combination and the parallel combination) are now connected in series with each other. Therefore, the total resistance $R_{\text{final}}$ of this final combination is:

$R_{\text{final}}=R_s+R_p=50+2=52\Omega$