Question

The length of a potentiometer wire is 'L'. A cell of e.m.f. 'E' is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$, then using the same cell null point is obtained at

• A. $\frac{L}{4}$
• B. $\frac{L}{3}$
• C. $\frac{L}{2}$
• D. $\frac{3L}{4}$

Hint:

Balancing length is directly proportional to the total length of the potentiometer wire for a fixed EMF.

Answer 1

The Correct Option is B

Step 1: Potentiometer Principle
$E = k \cdot l$, where $k$ is potential gradient ($V/L_{total}$) and $l$ is balancing length. Since the driver cell ($V$) and $E$ are constant, $E = \frac{V}{L_{total}} \cdot l$.
Step 2: Initial State
$E = \frac{V}{L} \cdot \frac{L}{4} = \frac{V}{4}$.
Step 3: Final State Calculation
New total length $L' = L + \frac{L}{3} = \frac{4L}{3}$.
New balance length $l'$: $E = \frac{V}{(4L/3)} \cdot l'$.
Step 4: Solving for l'
$\frac{V}{4} = \frac{3V}{4L} \cdot l' \Rightarrow \frac{1}{4} = \frac{3}{4L} \cdot l' \Rightarrow l' = \frac{L}{3}$.
Final Answer:(B)