The length of a potentiometer wire is $L$. A cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from…

The length of a potentiometer wire is $L$. A cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased $\frac{L}{2}$ at what distance will the same cell gives a balance point?
  1. $\frac{2 L}{3}$
  2. $\frac{L}{2}$
  3. $\frac{L}{6}$
  4. $\frac{4 L}{3}$

Solution

Let $x$ be the required balance length. Potential gradient in the first case $=\frac{E_0}{L}$ \(E=\left(\frac{1}{3}\right) \cdot\left(\frac{E_0}{L}\right)=\frac{E_0}{3}\) ...(1)
Potential gradient in the second case is given by,
\(\begin{aligned}
& \frac{E_0}{\left(\frac{3 L}{2}\right)}=\frac{2 E_0}{3 L} \\
& \therefore E=(x) \frac{2 E_0}{3 l} \quad \ldots (2)
\end{aligned}\) From equation (1) and (2), we get $\begin{aligned} & \frac{E_0}{3}=\left(\frac{2 E_0}{3 L}\right) x \\ & \Rightarrow x=\frac{L}{2}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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