Question

The length of a potentiometer wire is $l$. A cell 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) $\dfrac{4}{3}l$

Answer 1

Hint: When the length of wire is changed, its resistance increases and hence one needs to change the position of the jockey to balance the potentiometer.
By balancing the potentiometer, one means that potential difference between the wire and the jockey is equal to the emf of the cell whose emf one needs to find out.

Formula used:
$\dfrac{{{E_1}}}{{{l_1}}} = \dfrac{{{E_2}}}{{{l_2}}}$

Complete step by step solution:
In the above formula, for emf ${E_1}$ balance point is achieved at a length equal to ${l_1}$ .
For emf equal to ${E_2}$, the balance point will change. This new balance point will be at a length equal to ${l_2}$.
In this case, assume that the new length where the potentiometer will be balanced is at y. It is given that the length at which this potentiometer is balanced is $\dfrac{l}{3}$.
$\therefore $ Potential gradient is $E = \dfrac{l}{3} \times \dfrac{{{E_0}}}{l} = \dfrac{{{E_0}}}{3}$ ………...equation $(1)$
When the length of wire is increased, $l$ in the last equation needs to be replaced by $\dfrac{l}{3}$
$\therefore E = \dfrac{{{E_0}}}{{\dfrac{{3l}}{2}}} = \dfrac{{2{E_0}}}{{3l}} \times y$ ……...equation $(2)$
From $(1)\& (2)$ , $\dfrac{{{E_0}}}{3} = \dfrac{{2{E_0}}}{{3l}}y$
$ \Rightarrow y = \dfrac{l}{2}$
Therefore, the new length at which the balance point will be obtained is $\dfrac{l}{2}$ measured from the positive end of the wire.

Therefore, option B is correct.

Note: All lengths are to be measured from the positive end of wire to avoid confusion.
Galvanometer is connected so as to measure the amount of current flowing at a point.
Rheostat is connected to change the resistances.