d
As we know, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{5}{50}=0.1\)

\(I' = 0.099\)

When Galvanometer is connected

\(R_{e q}=50+\frac{100 S}{100+S}=\frac{V}{I}\)

\(\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=\frac{5}{0.099}-50\)

\(\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=50.50-50 \Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=0.5\)

\(\Rightarrow 100 \mathrm{S}=50+0.55 \Rightarrow 99.5 \mathrm{S}=50\)

\(S=\frac{50}{99.05}=0.5 \,\Omega\)

So, shunt of resistance \(=0.5 \,\Omega\) is onnected in parallel with the galvanometer.