Question

A potentiometer consists of a wire of length $4\mathrm{m}$ and resistance $10\mathrm{\Omega }$. It is connected to a cell of EMF $3\mathrm{V}$. The potential gradient of wire is

• A. $0.75\mathrm{V}/\mathrm{m}$
• B. $2\mathrm{V}/\mathrm{m}$
• C. $6\mathrm{V}/\mathrm{m}$
• D. $10\mathrm{V}/\mathrm{m}$

Answer 1

The correct option is A

$0.75\mathrm{V}/\mathrm{m}$

Potentiometer:

1. A potentiometer is an device or instrument which is used for measuring an electromotive force by balancing it against the potential difference obtained by passing a known current via a know variable resistance.

Potential gradient:

1. It is the ratio of potential over potentiometer to the length of the potentiometer.

Determine the potential gradient of the wire.

It is given that, potentiometer wire length $\mathrm{l}$ is $4\mathrm{m}$, resistance of the potentiometer, $\mathrm{R}$ is $10\mathrm{\Omega }$ and Electromagnetic field of the cell $\mathrm{E}$ is $3\mathrm{V}$.

We have to find the potential gradient of the wire.

The potential gradient $\mathrm{k}$ is,

$\mathrm{k}=\frac{\mathrm{V}}{\mathrm{L}}$

Here, $\mathrm{V}$ is the potential difference which is the same as the EMF of the cell.

Substitute the give values in the above formula, we get,

$\mathrm{k}=\frac{3\mathrm{V}}{4\mathrm{m}}$

$\mathrm{k}=0.75\mathrm{V}/\mathrm{m}$

So, the potential gradient of the wire is $0.75\mathrm{V}/\mathrm{m}$.

Hence , option $\mathrm{A}$ is correct answer.