A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2:3:4$ . Their lengths must be

The frequency produced by a stretched wire is given by
$f=\frac{p}{2l}\sqrt{\frac{T}{m}}$ … (i)
Where, p = number of loops formed in vibration,
$l$ = length of wire,
T = tension in wire,
and m = mass per unit length of wire
So, form Eq (i), we get
$f\propto \frac{1}{l}$
Here, the ratio of frequencies of three segments is
$f_1:f_2:f_3=2:3:4$
$\Rightarrow l_1:l_2:l_3=\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$
$=6:4:3$
The length of wire, L = 260 cm, so
$l_1=\frac{6}{13}\times 260=120$ cm
$l_2=\frac{4}{13}\times 260=80$ cm
$and{l}_3=\frac{3}{13}\times 260=60$ cm