The length of a potentiometer wire is `l`. A cell of emf `E` is balanced at a length l/3 from the positive end of the wire. If the length of the wire is increased by l/2. 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 balanc

1.	The length of a potentiometer wire is `l`. A cell of emf `E` is balanced at a length l/3 from the positive end of the wire. If the length of the wire is increased by l/2. At what distance will the same cell give a balance point.
Answer» Correct Answer - 5 Let `E_(0)` be the pot diff applied across the total length `l(= 10cm)` of potentiometer wire Potential gradient in the first case `= (E_(0))/(l)` As per question, `E = (l)/(3)((E_(0))/(l)) = (E_(0))/(3)`….(i) potential gradient in the second case `= (E_(0))/(3l//2) = (2E_(0))/(3l)` If `x` is the desired length of potentiometer to balance the emf `E` of the cell, then `E = x xx (2E_(0))/(3l)`......(iii) From (i) and (ii) we have `(E_(0))/(3) = x xx (2E_(0))/(3l)` or `x = (l)/(2)= (10)/(2) = 5 cm`

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The length of a potentiometer wire is `l`. A cell of emf `E` is balanced at a length l/3 from the positive end of the wire. If the length of the wire is increased by l/2. At what distance will the same cell give a balance point.

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