Initially neutral position is $60\, \mathrm{cm} .$ from $\mathrm{A}$,

so

$\frac{4}{60}=\frac{Q}{40} \Rightarrow Q=\frac{16}{6}=\frac{8}{3}\, \Omega$

Now, when unknown resistance $\mathrm{R}$ is connected in series to $P$, neutral point is $80 \,cm$ from Athen,

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

$\frac{4+R}{80}=\frac{8}{60}$

$R=\frac{64}{6}-4=\frac{64-24}{6}=\frac{40}{6} \,\Omega$

Hence, the value of unknown resistance $\mathrm{R}$

is $=\frac{20}{3} \,\Omega$