Question

The length of a potentiometer wires is $l$. A cell of $emf \,E$ is balanced at a length $\left(\frac{l}{3}\right)$ from positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$, the distance at which the same cell gives the balancing point is
(Cell in the primary is ideal and no series resistance is present in the primary circuit.)

• A. $\frac{21}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{6}$
• D. $\frac{41}{3}$

Answer 1

Answer: (B)

According to question, the figure is as

If $K$ be the potential gradient of the potentiometer
wire, then emf of the cell which gives balancing
length $\frac{1}{3}$ is given by
$E=K \cdot \frac{l}{3} \cdots (i)$
Where, $E=\frac{V}{l} \cdot \frac{l}{3}=\frac{V}{3} \cdots (ii) [\because K=\frac{V}{l}]$
When, length of potentiometer wire is increased 1
by $\frac{l}{2}$, then new length,
$l_{1}=l+\frac{l}{2}=\frac{3 l}{2}$
$\therefore $ New potential gradient,
$K^{\prime}=\frac{V}{\frac{3 l}{2}} \Rightarrow K^{\prime}=\frac{2 V}{3 l}$
If $l'$ be the new balancing length,
then, $E=K' l' .....(iii)$
or $E=\frac{2 V}{3 l} l'$
$ \frac{2 V}{3 l} \cdot l'=\frac{V}{3} \Rightarrow l'=\frac{l}{2} [\text { From Eq. (iii), } E=\frac{V}{3}]$