d
(d)

\(k=73.5 \,Nm ^{-1} \quad \text { Force }=5 \times 9.8\)

In figure \((1)\)

\(5 \times 9.8=(2 k) x_1\)

\(\therefore x_1=\frac{5 \times 9.8}{2 \times 73.5}=\frac{1}{3}\)

In figure \((2)\)

\(5 \times 9.8=\frac{k \times k}{k+k} \times x_2\)

or \(5 \times 9.8=\frac{k}{2} \times x_2\)

\(x_2=\frac{98}{73.5}=\frac{4}{3}\)

In figure \((3)\)

\(5 \times 9.8=k x_3\)

\(x_3=\frac{5 \times 9.8}{73.5}=\frac{2}{3}\)