A conductor is connected to a battery of 2" " V. Let V_1 be the potential difference across the conductor. When a battery of 4 V is connected, the potential difference becomes V_2. The ratio (V_1/V_2 ) is:

Correct option is D

Given:

A conductor is connected to a battery.

When the conductor is connected to a battery of 2 V, the potential difference across the conductor is V_1

When the conductor is connected to a battery of 4 V, the potential difference across the conductor becomes V_2​.

Solution:

​$$\text{Ohm's Law:} \quad V = I \cdot R$$

$$\text{Since } R \text{ is constant, } V \propto \text{Battery Voltage}$$ 

$$\text{Ratio:} \quad \frac{V_1}{V_2} = \frac{\text{Battery Voltage for } V_1}{\text{Battery Voltage for } V_2}$$

$$\frac{V_1}{V_2} = \frac{2}{4} = \frac{1}{2}$$

Thus the correct answer is (D)