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: Understanding the Question:
The principle of a potentiometer is that the EMF of a cell is proportional to its balancing length. The proportionality constant is the potential gradient $k$.
Step 2: Key Formula or Approach:
$E = k l = \left(\frac{V}{L}\right)l$, where $V$ is the potential difference across the total wire length $L$.
Step 3: Detailed Explanation:
Initial state:
Length $= L$. Potential gradient $k_1 = \frac{V}{L}$.
Balancing length $l_1 = \frac{L}{4}$.
EMF $E = k_1 l_1 = \left(\frac{V}{L}\right) \cdot \left(\frac{L}{4}\right) = \frac{V}{4}$.
Final state:
New length $L' = L + \frac{L}{3} = \frac{4L}{3}$.
New potential gradient $k_2 = \frac{V}{L'} = \frac{V}{4L/3} = \frac{3V}{4L}$.
Let the new balancing length be $l_2$. For the same cell:
\[ E = k_2 l_2 \]
\[ \frac{V}{4} = \left(\frac{3V}{4L}\right) \cdot l_2 \]
\[ 1 = \left(\frac{3}{L}\right) \cdot l_2 \implies l_2 = \frac{L}{3} \]
Step 4: Final Answer:
The new null point is obtained at $L/3$.