A potentiometer wire has length 4 m and resistance 8Ω. 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 1 mV per cm on the wire is

the total resistance after adding resistance R in series with potentiometer wire=8Ω+RΩ
the current in the circuit will be
i=V/Req=2/(R+8)
the potential drop across potentiometer wire will be
iRpotentiometer=2/(R+8)×8
the potential gradient along length 4 m is given as=
Vpotetiometer/4=2/(R+8)×8/4
the gradient is given as
1mV/cm=0.1V/m
2/(R+8)×8/4=0.1
R=32Ω