A standerd cell emf `1.08 V `is balance by the potential difference across `91 cm` of a meter long wire applied by a cell of emf `2 V` through a series resistor of resistance `2 Omega`. The internal resistance of the cell is zero. Find the resistance per unit length of the potentiometer wire.

Answer

` 0.30 Omega cm `

Explanation

Let `l` be the resistance of potentimeter wire of length `100 cm` current `l` in the potentimeter wire is
`I = ((epsilon )/(R + r)) = (2)/(2 + r)`
potential difference across `100 cm` potentimeter wire `I r = (2 xx r)/(2 + r)`
Potential difference across `91 cm` length is
`1.08 = (2r )/((2 + r)) = xx (91)/(100)`
On solving `r = 3 OMega `
resistance per unit length of potentiometer wire
`=(r )/(100) = (3)/(100) = 0.30 Omega cm `