Q 12-03-007NEETNEET 2024Top questionEasy
A wire of length '$l$' 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:
Answer: (D) $52\ \Omega$
Each part has resistance $\dfrac{100}{10} = 10\ \Omega$.
Five in series: $5 \times 10 = 50\ \Omega$.
Five in parallel: $\dfrac{10}{5} = 2\ \Omega$.
Total: $50 + 2 = 52\ \Omega$.
Solution by Sreeraj P, M.Sc Physics