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.

Key concepts

Null point

In the context of a potentiometer experiment with a Leclanche cell, the null point represents a position along the potentiometer wire where the voltage across the wire is balanced and equals the voltage across the unknown or test battery (the Leclanche cell, in this case). At this point, there is no current flowing through the galvanometer connected to the circuit. The galvanometer is essentially a device that detects current, and thus, a reading of zero (null) indicates a balance.

The null point is crucial because it allows for the precise measurement of the electromotive force (emf) of a cell without drawing any current from it. This is one of the key advantages of using a potentiometer over a voltmeter for precise measurements. In this exercise, the null point was initially measured at 1 m, meaning this is where the electrical potential (voltage) matched the emf of the Leclanche cell precisely.

Leclanche cell

A Leclanche cell is a type of primary cell or battery that is often used in school experiments due to its relatively simple and reliable chemical makeup. This cell is named after its inventor, Georges Leclanché. Typically, it consists of a manganese dioxide positive electrode and a zinc negative electrode, with an electrolyte of ammonium chloride. Unlike rechargeable batteries, primary cells like the Leclanche cell cannot be recharged once depleted.

In potentiometer experiments, the Leclanche cell is often used as it provides a steady voltage, which is essential for finding the null point accurately. It was historically popular in devices such as remote controls and torches but is less common today due to more advanced batteries. In the exercise, this cell was used to find the original null point and then observe changes when the potentiometer wire length was altered.

Length ratio

The concept of length ratio in this exercise is about understanding how the change in length of the potentiometer wire affects the null point position. When the length of the wire is increased, the potential difference along the wire changes proportionally.

The ratio is calculated by comparing the new length of the wire to its original length. In this exercise, the potentiometer wire was initially 4 m long and increased to 5 m. Therefore, the length ratio is calculated as \( \frac{5}{4} \). This ratio indicates that any position measured on the original wire must now be proportional to this ratio on the extended wire.

By multiplying the initial null point (1 m) by the length ratio, the new null point is determined. This method ensures that we account for the uniform distribution of voltage along the wire and find the new balanced position when the length is altered. The result was that the new null point shifted to 1.25 m on the extended wire.