A metal rod of length 10 cm and a rectangular cross-section \( 1\,\mathrm{cm} \times \displaystyle\frac{1}{2}\,\mathrm{cm} \) is connected to a battery across opposite faces. The resistance will be :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → maximum when the battery is connected across \( 1\,\mathrm{cm} \times \displaystyle\frac{1}{2}\,\mathrm{cm} \) faces.
The resistance of a conductor is \( R = \rho\,l/A \). When the battery is connected across two opposite faces, \( A \) is the area of those faces and \( l \) is the distance between them. The resistance is largest when the current has the longest path through the smallest area.
\( 1\,\mathrm{cm} \times 0.5\,\mathrm{cm}\ \text{faces}: l = 0.1\,\mathrm{m},\ A = 0.5\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle R_1 = \frac{\rho \times 0.1}{0.5\times10^{-4}} = 2000\,\rho \)
\( 10\,\mathrm{cm} \times 1\,\mathrm{cm}\ \text{faces}: l = 0.005\,\mathrm{m},\ A = 10\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle R_2 = \frac{\rho \times 0.005}{10\times10^{-4}} = 5\,\rho \)
\( 10\,\mathrm{cm} \times 0.5\,\mathrm{cm}\ \text{faces}: l = 0.01\,\mathrm{m},\ A = 5\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle R_3 = \frac{\rho \times 0.01}{5\times10^{-4}} = 20\,\rho \)
(\( \rho \) in \( \Omega\,\mathrm{m} \), \( R \) in \( \Omega \).) So the resistance is largest, \( 2000\,\rho \), across the \( 1\,\mathrm{cm} \times 0.5\,\mathrm{cm} \) end faces.