A potentiometer wire has length 4 m and resistance 8Ω . The...

$\dfrac{dV}{dt} = \dfrac{1mV}{cm}$
For a potentiometer $1cm$ of wire has $1mV$ of voltage
So,
$400cm$ has $0.4V .$
Given R must be connected in series
$\DeltaV = 0.4 = \dfrac{2}{8 + R} \times 8$
$8 + R = \dfrac{16}{0.4} = \dfrac{160}{4} = 40$
$⇒ R = 32Ω$