d
\(i_0 R_0=i_{100} R_{100} \quad\) [For same source]

\(\Rightarrow 2 R_0=1.2 R_0[1+100 \alpha]\)

\(\Rightarrow 1+100 \alpha=\frac{5}{3} \Rightarrow 100 \alpha=\frac{2}{3}\)

\(\Rightarrow 50 \alpha=\frac{1}{3}\)

\(\therefore i_{50} R_{50}=i_0 R_0\)

\(\Rightarrow i_{50}=\frac{i_0 R_0}{R_{50}}=\frac{2 \times R_0}{R_0(1+50 \alpha)}=\frac{2}{1+\frac{1}{3}}=1.5\,A\)

\(=15 \times 10^2\,mA\)