Q.
A conductor with rectangular cross-section has dimensions $(a \times 2a \times 4a)$ as shown in figure. Resistance across $AB$ is $R_{1}$, across $CD$ is $R_{2}$ and across $EF$ is $R_{3}.$Then
A) $R_{1} = R_{2} = R_{3}$
B) $R_{1} > R_{2} > R_{3}$
C) $R_{2} > R_{3} > R_{1}$
D) $R_{1} > R_{3} > R_{2}$
Explanation
(d) $R = \frac{\rho l}{A}.So\ R_{1} = \frac{\rho(4a)}{(2a)(a)} = \frac{2\rho}{a}$
$R_{2} = \frac{\rho(a)}{(4a)(2a)} = \frac{\rho}{8a}and\ R_{3} = \frac{\rho(2a)}{(4a)(a)} = \frac{\rho}{2a}$
$\therefore R_{1} > R_{3} > R_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.