b
Mass $(m)=5 kg$ and
time period $(T)=2 \pi$ sec.
Therefore time period $T=2 \pi \times \sqrt{\frac{m}{k}} \Rightarrow \sqrt{\frac{5}{k}}=1$
or $k=5 N / m$.
According to Hooke's Law, $F=-k l$.
Therefore decrease in length $(l)=-\frac{F}{k}=-\frac{5 g}{5}$ $=-g$ metres