A potentiometer wire of length 300cm is connected in series with a...

Let resistance of potentiometers wire is $R$, then the current through $AB$

${i}_{AB} = \dfrac{4}{R + 780}$

Potential difference across $AB$

${V}_{AB} = \dfrac{4R}{R + 780}$

Potential difference across $AC$

${V}_{AC} = {V}_{AB}\dfrac{l}{L} = \dfrac{4R}{\left(R + 780\right)} \times \dfrac{60}{300} = \dfrac{4R}{5\left(R + 780\right)}$

This should be equal to $20mV$

$⇒ \dfrac{4R}{5\left(R + 780\right)} = 20 \times {10}^{- 3} = 2 \times {10}^{- 2}$

$⇒ 4R = {10}^{- 1}\left(R + 780\right)$

$⇒ 4R - \dfrac{R}{10} = 78$

$⇒ \dfrac{39R}{10} = 78$

$⇒ R = 20Ω$