Q109 / 180 · JIPMER 2015 · Physics
A copper rod of length 20 cm and cross-sectional area $2 \mathrm{~mm}^2$ is joined with a similar aluminium rod as shown below
The resistance of pair of rods is $\left(\rho_{\mathrm{Al}}=2.6 \times 10^{-8} \quad \Omega-\mathrm{m}\right.$ and $\quad \rho_{\mathrm{Cu}}=1.7 \times 10^{-8}$ $\Omega-\mathrm{m})$
The resistance of pair of rods is $\left(\rho_{\mathrm{Al}}=2.6 \times 10^{-8} \quad \Omega-\mathrm{m}\right.$ and $\quad \rho_{\mathrm{Cu}}=1.7 \times 10^{-8}$ $\Omega-\mathrm{m})$- Correct answerA$1.0 \mathrm{~m} \Omega$
- B$2.0 \mathrm{~m} \Omega$
- C$3.0 \mathrm{~m} \Omega$
- DNone of these
Solution
The resistance of the copper rod $=\rho \frac{1}{a}$
$=\frac{1.7 \times 10^{-8} \times 20 \times 10^{-2}}{2.0 \times 10^{-6}}=1.7 \times 10^{-3} \Omega$
Similarly the resistance of aluminium rod
$=2.6 \times 10^{-3} \Omega$
The equivalent resistance
$\begin{aligned} & \frac{1}{R}=\frac{1}{1.7 \times 10^{-3}}+\frac{1}{2.6 \times 10^{-3}} \\ & R=\frac{1.7 \times 2.6}{4.3} \times 10^{-3}=1.0 \mathrm{~m} \Omega\end{aligned}$
Hence, the correct answer is option: 'A'.