Q 12-03-203JEE MainJEE Main 2021 (22 Jul, Shift 1)Medium
A Copper (Cu) rod of length 25 cm and cross-sectional area $3\ \text{mm}^2$ is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends $A$ and $B$.
(Take resistivity of Copper $= 1.7\times10^{-8}\ \Omega$ m, Resistivity of aluminium $= 2.6\times10^{-8}\ \Omega$ m)
Answer: (D) 0.858 m$\Omega$
$R_{Cu} = \dfrac{1.7\times10^{-8}\times0.25}{3\times10^{-6}} = 1.417\ \text{m}\Omega$ and $R_{Al} = \dfrac{2.6\times10^{-8}\times0.25}{3\times10^{-6}} = 2.167\ \text{m}\Omega$.
In parallel:
$$R = \frac{1.417\times2.167}{1.417 + 2.167} \approx 0.858\ \text{m}\Omega$$
Solution by Sreeraj P, M.Sc Physics