Question

The current-voltage graphs for a given conducting sample at two different temperatures \(T_1\) and \(T_2\) are shown in the figure below. \(R_1\) is the resistance of the sample at temperature \(T_1\) and \(R_2\) is the resistance at temperature \(T_2\). Choose the correct answer.

• A. \(R_2\gt R_1,\; T_2\gt T_1\)
• B. \(R_2\gt R_1,\; T_2\lt T_1\)
• C. \(R_2\lt R_1,\; T_2\gt T_1\)
• D. \(R_2\lt R_1,\; T_2\lt T_1\)

Hint:

In an \(I\)-\(V\) graph, \[ \text{slope}=\frac{1}{R}. \] For metallic conductors, \[ R \propto T. \] So a smaller slope corresponds to a higher temperature.

Answer 1

The Correct Option is B

Concept: For an ohmic conductor, \[ V=IR \] or \[ I=\frac{V}{R}. \] Hence, in an \(I\)-\(V\) graph, \[ \text{slope}=\frac{I}{V}=\frac{1}{R}. \] Greater slope means smaller resistance.

Step 1: Compare the slopes. The line corresponding to \(T_1\) is steeper than the line corresponding to \(T_2\). Therefore, \[ \frac{1}{R_1}\gt \frac{1}{R_2}. \] Hence, \[ R_2\gt R_1. \]

Step 2: Relate resistance and temperature. For a metallic conductor, \[ R \uparrow \quad \text{as} \quad T \uparrow. \] Since \[ R_2\gt R_1, \] it follows that \[ T_2\gt T_1. \] \[\begin{aligned} \boxed{R_2\gt R_1,\quad T_2\gt T_1} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.

Note: The marked answer in the image appears inconsistent with the physics of metallic conductors. If the sample is a normal conductor (as stated), option \(\mathbf{(A)}\) is the correct choice.