KCET · Physics · Current Electricity
A metal rod of length \(10 \mathrm{~cm}\) and a rectangular cross-section of \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\) is connected to a battery across opposite faces. The resistance will be
- A maximum when the battery is connected across \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\) faces
- B maximum when the battery is connected across \(10 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\) faces
- C maximum when the battery is connected across \(10 \mathrm{~cm} \times 1 \mathrm{~cm}\) faces
- D same irrespective of the three faces
Answer & Solution
Correct Answer
(A) maximum when the battery is connected across \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\) faces
Step-by-step Solution
Detailed explanation
Given, length, \(l=10 \mathrm{~cm}\)
Resistance of metal rod.
\(R=\rho \cdot \frac{l}{A}\)
i.e., \(\quad R \propto \frac{1}{A}\)
Since, among given faces in the options, area is minimum corresponding to face \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\). Hence, resistance will be maximum when the battery is connected across the face of dimension \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\).
Resistance of metal rod.
\(R=\rho \cdot \frac{l}{A}\)
i.e., \(\quad R \propto \frac{1}{A}\)
Since, among given faces in the options, area is minimum corresponding to face \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\). Hence, resistance will be maximum when the battery is connected across the face of dimension \(1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}\).
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