JEE Mains · Physics · STD 12 - 3. current electricity
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 \(=2.6 \times 10^{-8}\, \Omega \,{m}\) )

- A \(1.420\)
- B \(0.0858\)
- C \(2.170\)
- D \(0.858\)
Answer & Solution
Correct Answer
(D) \(0.858\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) \(R=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{\ell}{A} \cdot \frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}\) \(R=\frac{25 \times 10^{-2}}{3 \times 10^{-6}} \times \frac{1.7 \times 2.6 \times 10^{-16}}{4.3 \times 10^{-8}}\)…
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