• Step 1: Calculate current in the primary circuit:
  $$I = \frac{E_1}{R_{AB} + r_1} = \frac{2.4\,\text{V}}{10\,\Omega + 2\,\Omega} = \frac{2.4}{12} = 0.2\,\text{A}$$
• Step 2: Potential drop across the potentiometer wire AB ($L = 2\,\text{m}$):
  $$V_{AB} = I \times R_{AB} = 0.2\,\text{A} \times 10\,\Omega = 2.0\,\text{V}$$
• Step 3: Potential gradient along the wire ($k$):
  $$k = \frac{V_{AB}}{L} = \frac{2.0\,\text{V}}{2.0\,\text{m}} = 1.0\,\text{V/m}$$
• Step 4: At balance point ($l = 1.6\,\text{m}$), no current is drawn from the standard cell, so its internal resistance ($0.2\,\Omega$) produces zero voltage drop:
  $$V = k \times l = 1.0\,\text{V/m} \times 1.6\,\text{m} = 1.6\,\text{V}$$

The drift velocity of conduction electrons in a conductor of length $l$ under an applied potential difference $V$ is given by:

where:

• $e$ is electron charge
• $m$ is electron mass
• $\tau$ is relaxation time
• $V$ is applied voltage
• $l$ is conductor length

For a given conductor at constant temperature, $v_d \propto V$. Hence, option 2 is the correct answer.

In semiconductors, the charge carrier density $n$ increases exponentially with temperature:

Since resistivity is inversely proportional to carrier density:

As temperature $T$ increases, resistivity $\rho$ decreases rapidly (exponential decay).

Graph 3 correctly represents this non-linear exponential decrease of resistivity with temperature.