Two potentiometer wires `w_(1)` and `w_(2)` of equal length `l` connected to a battery of emf `epsilon_(P)` and internal resistance 'r' as shown through two switches `s_(1)` and `s_(2)`. A battery of emf `epsilon` is balanced on these potentiometer wires. if potentiometer wire `w_(1)` is of resistance 2r and balancing length on `w_(1)` is `l//2` when only `s_(1)` is closed and `s_(2)` is open. On closing `s_(2)` and opening `s_(1)` the balancing length on `w_(2)` is found to be `((2l)/(3))` then find the resistance of potentiometer wire `w_(2)`

Explanation

For `w_(1),varepsilon =1/2[(varepsilon_(p))/(1+2)2/l]`
For `w_(2), varepsilon = (2l)/3[(varepsilon_(p))/(1+R) R/l]`
Dividing eq (1) by (2) and on solving we get Resistance of wire `w_(2)=1 Omega`