Q87 / 120 · JEE Advanced 2006 · Physics
Two bars of radius ' $r$ ' and ' $2 r$ ' are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of the following is correct?
- Correct answerAHeat produced in bar $B C$ is 4 times the heat produced in bar $A B$
- BElectric field in both halves is equal
- CCurrent density across $A B$ is doubled that of across $B C$
- DPotential difference across $A B$ is 4 times that of across $B C$
Solution
Current flowing through both the bars is equal. Now, the heat produced is given by
$$
\begin{aligned}
H & =I^2 R t \text { or } H \propto R \text { or } \frac{H_{A B}}{H_{B C}}=\frac{R_{A B}}{R_{B C}} \\
& =\frac{\left(\frac{1}{2 r}\right)^2}{\left(\frac{1}{r}\right)^2} \\
& =\frac{1}{4} \text { or } H_{B C}=4 H_{A B}
\end{aligned}
$$
$$
\left(\text { as } R \propto \frac{1}{A} \propto \frac{1}{r^2}\right)
$$
Hence, the correct answer is option: 'A'.