Question

The length of potentiometer wire is $l$. A call of emf $E$ is balanced at length $\dfrac{l}{3}$ from the positive end of the wire. If the length of the wire is increased by $\dfrac{l}{2}$. Then, at what distance will the same cell give a balance point?
A. $\dfrac{2l}{3}$
B. $\dfrac{l}{2}$
C. $\dfrac{l}{6}$
D. $\dfrac{4l}{3}$

Answer 1

Hint: Generally, when the arrangement is not balanced, there is some value of potential difference across the galvanometer. At balance point, this value becomes zero. We can say that the potential difference of wire between the end and jockey becomes equal to the emf of the experimental cell when balance point is obtained.

Formula used:
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{{{l}_{1}}}{{{l}_{2}}}$

Complete step by step answer:
In a potentiometer, a balance point is obtained when the potential difference of wire between the end and jockey becomes equal to the emf of the experimental cell. At balance point, the value of potential difference across the galvanometer becomes zero.
We calculate the balance point along the slide wire where there is zero deflection in the galvanometer.
When cell has emf ${{E}_{1}}$, balance point is obtained at ${{l}_{1}}$
When cell has emf ${{E}_{2}}$, balance point is obtained at ${{l}_{2}}$
By the equation of balance point in potentiometer we have,
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{{{l}_{1}}}{{{l}_{2}}}$
Let us take the required length of wire as $x$, where the cell will give a balance point
Potential gradient in the case where balance point is at $\dfrac{l}{3}$ will be equal to $E=\left( \dfrac{l}{3} \right)\times \left( \dfrac{{{E}_{o}}}{l} \right)=\dfrac{{{E}_{o}}}{3}$
Potential gradient in the case when length of wire is increased by $\dfrac{l}{2}$ will be equal to
\[E=\dfrac{{{E}_{o}}}{\dfrac{3l}{2}}=\left( \dfrac{2{{E}_{o}}}{3l} \right)\times x\]
Comparing above equations:
$\begin{align}
  & \dfrac{{{E}_{o}}}{3}=\left( \dfrac{2{{E}_{o}}}{3l} \right)\times x \\
 & x=\dfrac{l}{2} \\
\end{align}$
Balance point will now be obtained at a length $\dfrac{l}{2}$ from the positive end of the wire.
Hence, the correct option is B.

Note: Students should keep in mind that we would take the balance point length from the positive end of the wire; this would ensure no calculation error while finding the balance point.