Question

The length of a potentiometer wire is \(L\). A cell of e.m.f. \(E\) is balanced at a length \( \frac{L}{5} \) from the positive end of the wire. If the length of the wire is increased by \( \frac{L}{2} \), at what distance will the same cell give a balance point?

• A. \( \dfrac{5L}{12} \)
• B. \( \dfrac{2L}{15} \)
• C. \( \dfrac{4L}{15} \)
• D. \( \dfrac{3L}{10} \)

Hint:

In potentiometer problems, emf remains constant while potential gradient changes with wire length.

Answer 1

The Correct Option is D

Step 1: Use potentiometer principle.
The emf of the cell is proportional to the balancing length: \[ E \propto l \]

Step 2: Write relation for original wire.
\[ E = k \cdot \frac{L}{5} \]

Step 3: Find new potential gradient.
New length of wire \[ L' = L + \frac{L}{2} = \frac{3L}{2} \] New potential gradient \[ k' = \frac{kL}{L'} = \frac{2k}{3} \]

Step 4: Find new balancing length.
\[ E = k' l' \Rightarrow k \frac{L}{5} = \frac{2k}{3} l' \] \[ l' = \frac{3L}{10} \]