Question

In the experiment of calibration of a voltmeter, a standard cell of $emf=1.1V$ is balanced against a $440cm$ of potentiometer wire. The potential difference across the ends of resistance is found to balance against $220cm$ of wire. The corresponding error in the reading of the voltmeter will be:
$A)-0.15V$
$B)0.15V$
$C)0.5V$
$D)-0.05V$

Answer 1

A potentiometer is a measuring device, which is used to measure the potential difference in an electric circuit. A potentiometer is constructed using a long wire with a uniform area of cross-section. A voltmeter can also be used to measure the potential difference in a circuit. But, when we use a voltmeter directly, there are chances of error in the reading of the voltmeter, because the voltmeter draws a small amount of current from the voltage source in the circuit. This error in voltage reading can be rectified by using a potentiometer, which determines the accurate value of the potential difference in a circuit.

Complete step-by-step solution:
In the question, we are given that a standard cell of $emf=1.1V$ is balanced against $440cm$ of potentiometer wire. We know that the potential difference across a cell is proportional to the balancing length in a potentiometer. Mathematically, this can be represented as
$V\propto L\Rightarrow V=KL$
where $K$ is the constant of proportionality or the potential gradient of the wire. Potential gradient refers to the decrease in potential per unit length. It is given by
$K=\dfrac{V}{L}$
The potential gradient of the wire used in the above setup can be calculated as follows.
$K=\dfrac{V}{L}=\dfrac{1.1V}{440cm}=0.0025Vc{{m}^{-1}}$
Now, at a distance of $220cm$ in the wire, the potential difference is given by
$V=KL=0.0025Vc{{m}^{-1}}\times 220cm=0.55V$
This potential difference measured using a potentiometer is accurate because potentiometers do not draw current from the voltage source of the circuit.
If directly, a voltmeter was used to measure this potential difference, it would have recorded $0.5V$ or $0.6V$, because the least count of the voltage source is $0.1V$. This can be assumed because $0.55V$ lies in between $0.5V$ and $0.6V$.
If the voltmeter had recorded $0.5V$, the error in potential difference would be equal to
$0.5V-0.55V=-0.05V$
If the voltmeter had recorded $0.6V$, the error in potential difference would be equal to
$0.6V-0.55V=0.05V$
The only option which matches the above calculations is $D$. Hence, option $D$ is the correct answer.

Additional Information:
A potentiometer is constructed using a long wire of particular length and uniform area of cross-section. This circuit consists of a battery, a key, and a rheostat. It also consists of a galvanometer to record the amount of current flow in the circuit, as well as a jockey, to record the distance at which the voltage source is balanced or there is no current flow in the circuit.

Note: A potentiometer can also be used to compare the $emf's$ of two cells. This can be explained using the relation:
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{{{L}_{1}}}{{{L}_{2}}}$
where
${{E}_{1}}$ and ${{E}_{2}}$ are the $emf's$ of the cells to be compared
${{L}_{1}}$ and ${{L}_{2}}$ are the corresponding balancing lengths
Note that the potential gradient cancels out because the same wire of particular length and uniform cross-section is used in the potentiometer.