Solve this following
Question:

A Copper $(\mathrm{Cu})$ rod of length $25 \mathrm{~cm}$ and crosssectional area $3 \mathrm{~mm}^{2}$ is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends $\mathrm{A}$ and $B$.

(Take Resistivity of Copper $=1.7 \times 10^{-8} \Omega \mathrm{m}$ Resistivity of Aluminium $=2.6 \times 10^{-8} \Omega \mathrm{m}$ )

1. $2.170 \mathrm{~m} \Omega$

2. $1.420 \mathrm{~m} \Omega$

3. $0.0858 \mathrm{~m} \Omega$

4. $0.858 \mathrm{~m} \Omega$

Correct Option: , 4

Solution:

$\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{\ell}{\mathrm{A}} \cdot \frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$

$R=\frac{25 \times 10^{-2}}{3 \times 10^{-6}} \times \frac{1.7 \times 2.6 \times 10^{-16}}{4.3 \times 10^{-8}}$

$\mathrm{R}=0.858 \mathrm{~m} \Omega$