d
In balance position of bridge, \(\frac{P}{Q}=\frac{l}{(100-l)}\)

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 \(=\frac{20}{3}\, \Omega\)