The length of a wire between the two ends of a sonometer is 105 cm. Where the two bridges should be placed so that the fundamental frequencies of the three segments are in the ratio of 1 : 3 : 15.

From the law of length of stretched string, we have  $n_1{\mathcal{l}}_1=n_2{\mathcal{l}}_2=n_3{\mathcal{l}}_3$
Here  
$\therefore \frac{{\mathcal{l}}_1}{{\mathcal{l}}_2}=\frac{n_1}{n_2}=\frac{3}{1}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathcal{l}}_1}{{\mathcal{l}}_3}=\frac{n_3}{n_1}=15\text{/}1$
${\mathcal{l}}_2=\frac{{\mathcal{l}}_1}{3}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{\mathcal{l}}_3=\frac{{\mathcal{l}}_1}{15}$
The total length of the wire is 105 cm.
Therefore  ${\mathcal{l}}_1+{\mathcal{l}}_2+{\mathcal{l}}_3=105$
or   ${\mathcal{l}}_1+\frac{{\mathcal{l}}_1}{3}+\frac{{\mathcal{l}}_1}{15}=105$ or  $\frac{21{\mathcal{l}}_1}{15}=105$
${\mathcal{l}}_1=\frac{105\times 15}{21}=75\text{\hspace{0.17em}\hspace{0.17em}cm.}$
$\therefore {\mathcal{l}}_2=\frac{{\mathcal{l}}_1}{3}=\frac{75}{3}=25\text{\hspace{0.17em}\hspace{0.17em}cm}$
${\mathcal{l}}_3=\frac{{\mathcal{l}}_1}{15}=\frac{75}{15}=5\text{\hspace{0.17em}\hspace{0.17em}cm}$
Hence the bridge should be placed at 75 cm and (75 + 25) = 100 cm from one end.