Question:
The length of a potentiometer wire is \(l\). A cell of emf \(E\) is balanced at length \(l/3\) from the positive end of the wire. If the length of the wire is increased by \(l/2\), then at what distance will the same cell give a balance point?
The length of a potentiometer wire is \(l\). A cell of emf \(E\) is balanced at length \(l/3\) from the positive end of the wire. If the length of the wire is increased by \(l/2\), then at what distance will the same cell give a balance point?
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In a potentiometer, the balance length is proportional to the total wire length when the driving EMF and resistance per unit length remain constant.
Updated On: Apr 20, 2026
- \(\dfrac{2l}{3}\)
- \(\dfrac{l}{2}\)
- \(\dfrac{l}{6}\)
- \(\dfrac{4l}{3}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Potential gradient of a potentiometer: \(k = \frac{E_0}{L}\). It changes when the length of the wire changes (keeping \(E_0\) same).
Step 2: Detailed Explanation:
Original length: \(L_1 = l\), \(k_1 = \frac{E_0}{l}\). Balance point initially: \(x_1 = \frac{l}{3}\). New length: \(L_2 = l + \frac{l}{2} = \frac{3l}{2}\). New potential gradient: \(k_2 = \frac{E_0}{3l/2} = \frac{2E_0}{3l}\). Using \(E = k \cdot x\): \(x = \frac{E}{k_2} = \frac{k_1 \cdot (l/3)}{k_2}\). Substituting: \(x = \frac{(E_0/l)(l/3)}{2E_0/(3l)} = \frac{l}{2}\).
Step 3: Final Answer:
\[ \boxed{\frac{l}{2}} \]
Potential gradient of a potentiometer: \(k = \frac{E_0}{L}\). It changes when the length of the wire changes (keeping \(E_0\) same).
Step 2: Detailed Explanation:
Original length: \(L_1 = l\), \(k_1 = \frac{E_0}{l}\). Balance point initially: \(x_1 = \frac{l}{3}\). New length: \(L_2 = l + \frac{l}{2} = \frac{3l}{2}\). New potential gradient: \(k_2 = \frac{E_0}{3l/2} = \frac{2E_0}{3l}\). Using \(E = k \cdot x\): \(x = \frac{E}{k_2} = \frac{k_1 \cdot (l/3)}{k_2}\). Substituting: \(x = \frac{(E_0/l)(l/3)}{2E_0/(3l)} = \frac{l}{2}\).
Step 3: Final Answer:
\[ \boxed{\frac{l}{2}} \]
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