A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is

Required potential gradient =1mVcm-1=110Vm-1 Length of potentiometer wire, l=4 mSo potential difference across potentiometer wire =110×4=0.4 V                 . . . . .(i) In the circuit, potential difference across 8Ω=I×8=28+R×8                     . . . . (ii)Using equation (i) and (ii), we get 0.4=28+R×8410=168+R,8+R=40∴  R=32Ω