Question

A potentiometer wire, which is \(4 \mathrm{~m}\) long is connected to the terminals of a battery of steady voltage. A leclanche cell gives a null point at \(1 \mathrm{~m}\) if the length of the potentiometer wire be increased by $1 \mathrm{~m}$, What is the new position of the null point? (A) \(1.25 \mathrm{~m}\) (B) \(1.4 \mathrm{~m}\) (C) \(1.75 \mathrm{~m}\) (D) \(1.2 \mathrm{~m}\)

Short answer

The new position of the null point is \(1.25\mathrm{~m}\).

Step-by-step solution

Determine the ratio of the length increase

First, we need to identify the ratio of increase that happens when the length of the potentiometer wire is increased by \(1\mathrm{~m}\). The wire has an initial length of \(4\mathrm{~m}\) and is increased to \(4 + 1 = 5\mathrm{~m}\). Thus, its length ratio becomes: \[ \text{Length Ratio} = \frac{\text{New Length}}{\text{Initial Length}} = \frac{5}{4} \]

Calculate the new null point position

Now we need to find the new position of the null point. Since the potential difference along the potentiometer wire is directly proportional to its length, the same length ratio applies to the null point position. The leclanche cell gives an initial null point at \(1\mathrm{~m}\). So, we multiply this initial null point position by the length ratio found in the previous step to get the new null point position: \[ \text{New Null Point Position} = \text{Initial Null Point Position} \times \text{Length Ratio} \] \[ \text{New Null Point Position} = 1\mathrm{~m} \times \frac{5}{4} \] \[ \text{New Null Point Position} = 1.25\mathrm{~m} \] Therefore, the new position of the null point is \(1.25\mathrm{~m}\), which corresponds to the choice (A) in the given exercise.