Q. 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;
A) $55\,\Omega$
B) $60\,\Omega$
C) $26\,\Omega$
D) $52\,\Omega$
Explanation
[Solution by Claude - not from source] Each part is $10\,\Omega$. Five in series $=50\,\Omega$; five in parallel $=2\,\Omega$. Total $=52\,\Omega$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.