JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2021
- Hard
- MCQ
A Copper \((Cu)\) rod of length \(25\, {cm}\) and cross- sectional area \(3\, {mm}^{2}\) is joined with a similar Aluminium \((Al)\) rod as shown in figure. Find the resistance of the combination between the ends \(A\) and \(B\) (in \({m} \Omega\)) (Take Resistivity of Copper \(=1.7 \times 10^{-8}\, \Omega \,{m}\), Resistivity of Aluminium \(=2.6 \times 10^{-8}\, \Omega \,{m}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.858\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)
\(R=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{\ell}{A} \cdot \frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}\)
\(R=\frac{25 \times 10^{-2}}{3 \times 10^{-6}} \times \frac{1.7 \times 2.6 \times 10^{-16}}{4.3 \times 10^{-8}}\)
\(R=0.858\, {m}\, \Omega\)
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