d
Total length of the wire, \(L=114 \mathrm{cm}\)

\(\mathrm{n}_{1}: \mathrm{n}_{2}: \mathrm{n}_{3}=1: 3: 4\)

Let \(L_{1}, L_{2}\) and \(L_{3}\) be the lengths of the three parts

As \(n \propto \frac{1}{L}\)

\(\therefore \quad \mathrm{L}_{1}: \mathrm{L}_{2}: \mathrm{L}_{3}=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3\)

\(\therefore \quad \mathrm{L}_{1}=72 \mathrm{cm}\left(\frac{12}{12+4+3} \times 114\right)\)

\(\mathrm{L}_{2}=24 \mathrm{cm}\left(\frac{4}{19} \times 114\right)\)

and \(\mathrm{L}_{3}=18 \mathrm{cm}\left(\frac{3}{19} \times 114\right)\)

Hence the bridges should be placed at

\(72 \mathrm{cm}\) and \(72+24=96 \mathrm{cm}\) from oneend