Question

A potentiometer wire has length \(L\). For a given cell of e.m.f. \(E\), the balancing length is \(\frac{L}{3}\) from the positive end of the wire. If the length of potentiometer wire is increased by \(50\%\) then for the same cell the balance point is obtained at length (from positive end)

• A. \(\frac{L}{2}\)
• B. \(\frac{L}{5}\)
• C. \(\frac{L}{3}\)
• D. \(\frac{L}{4}\)

Hint:

The potential gradient falls when the same driver voltage is spread over a longer wire, so the balance length rises.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The balancing length gives \(E=k\,l\), where \(k=\dfrac{V}{L}\) is the potential gradient for the driving voltage \(V\).

Step 2: Original Case
\(E=\dfrac{V}{L}\cdot\dfrac{L}{3}=\dfrac{V}{3}\)

Step 3: New Wire
The new length is \(1.5L\), so the gradient is \(\dfrac{V}{1.5L}\) and \(E=\dfrac{V}{1.5L}\,l'\).
\(\dfrac{V}{3}=\dfrac{V\,l'}{1.5L}\Rightarrow l'=\dfrac{1.5L}{3}=\dfrac{L}{2}\)

Step 4: Compare the options
The balance point is at \(\dfrac{L}{2}\), option (A).

Final Answer:
The balance point is at a length of L/2 from the positive end. \[ \boxed{\text{(A) }\dfrac{L}{2}} \]