Sankalp NEET Full Test-1 Question-18 Solution

Question 18: A wire of length ‘ \ell ‘ and resistance 100 \Omega 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:(1) 26 \Omega

(2) 52 \Omega

(3) 55 \Omega

(4) 60 \Omega

Answer: Option (2)

Explanation:

The total resistance of the wire is 100 \Omega and it is divided into 10 equal parts,

so the resistance of each part is

\frac{100}{10}=10 \Omega

The first 5 parts are connected in series, so their equivalent resistance is

R_{\text{series}}=5 \times 10 = 50 \Omega

The next 5 parts are connected in parallel. For equal resistances in parallel,

the equivalent resistance is given by

R_{\text{parallel}}=\frac{R}{n}

Here, R=10 \Omega and n=5, so

R_{\text{parallel}}=\frac{10}{5}=2 \Omega

Finally, the series combination and the parallel combination are connected in series.

Therefore, the total resistance is

R_{\text{total}}=50+2=52 \Omega

Hence, the resistance of the final combination is 52 \Omega.