Q. Two bars of radius $r\ and\ 2r$ are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of following is correct?
A) Heat produced in bar $BC$ is 4 times the heat produced in bar $AB$
B) Electric field in both halves is equal
C) Current density across $AB$ is double that of across $BC$
D) Potential difference across $AB$ is 4 times that of across $BC$
Explanation
(a) Current flowing through both the bars is equal.
Now, the heat produced is given by
$H = I^{2}Rt$
$H \propto R\ or\frac{H_{AB}}{H_{BC}} = \frac{R_{AB}}{R_{BC}}$
$\ \ \ \ = \frac{(1/2r)^{2}}{(1/r)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( as\ R \propto \frac{1}{A} \propto \frac{1}{r^{2}} \right)$
$\ \ \ = \frac{1}{4}\ or\ H_{BC} = 4H_{AB}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.