Initially, the resistance P = 4 $$\Omega$$ and the neutral point N is at 60 cm from A. Therefore,

$${4 \over {60}} = {Q \over {40}} \Rightarrow Q = {{160} \over {60}} = \left( {{8 \over 3}} \right)\Omega $$

When an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A, we have

$${{4 + R} \over {80}} = {Q \over {20}}$$

$$ \Rightarrow {{4 + R} \over {80}} = {8 \over {60}} \Rightarrow 4 + R = {{64} \over 6}$$

$$ \Rightarrow R = {{64} \over 6} - 4 = {{64 - 24} \over 6} = {{40} \over 6}$$

$$ \Rightarrow R = \left( {{{20} \over 3}} \right)\Omega $$