A `5 m` potentiometer wire having `3 Omega` resistance per meter is connected to a storage cell of steady emf `2 V` and internal resistance `1 Omega`. A primary cell is balanced against `3.5 m` of it. When a resistance of `32//n Omega` is put in series with the storage cell, the null point shifts to the centre of the last wire, i.e., `4.5 m`. What is 'n'?

Answer

`(7)`

Explanation

This means you will not get the balancing point.
`I = (2)/(1 + 15) = (1)/(8)`, `E_(P) = (1)/(8) xx 3xx (7)/(2) = (21)/(16)`
`(21)/(16) = I xx 3 xx (9)/(2)` or `I = (7)/(72) A`
`(7)/(72) = (2)/(16 + R)` or `R = (32)/(7) Omega`