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…

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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.)

Aicharya · Accepted Answer

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 $\frac{l}{3}$ is given by $E=K \cdot \frac{l}{3}$...(i) Where, $\quad E=\frac{V}{l} \cdot \frac{l}{3}=\frac{V}{3} \ldots$ (ii) $\left[\because K=\frac{V}{l}\right]$ When, length of potentiometer wire is increased 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 $l^{\prime}$ be the new balancing length, $\begin{gathered} \text{then, } E=K^{\prime} l^{\prime} \quad \ldots (iii) \\ \text{or } E=\frac{2 V}{3 l} l^{\prime} \\ \frac{2 V}{3 l} \cdot l^{\prime}=\frac{V}{3} \Rightarrow l^{\prime}=\frac{l}{2} \quad\left[\text { From Eq. (iiii) } E=\frac{V}{3}\right] \end{gathered}$

The length of a potentiometer wires is l. A cell of emf \(E\) is balanced at a length…

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