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

Options

A. 80 cm, 60 cm, 120 cm
B. 120 cm, 80 cm, 60 cm
C. 60 cm, 80 cm, 120 cm
D. 120 cm, 60 cm, 80 cm

Answer

B

Explanation

The frequency produced by a streched wire is given by
`f = (p)/(2l) = sqrt((T)/(m)) " "……(i)`
where p = number of loops formed in vibrations,
l = length of wire
T = tension in wire ,
and m = mass per unit lenght of wire
So, form Eq, (i) we get
`f prop (1)/(l)`
Hence , the ratio of frequencies of three segment is
`f_(1) : f_(2): f_(3) = 2 : 3:4`
`rArr " "l_(1) : l_(2): l_(2) =(1)/(2): (1)/(3):(1)/(4)`
` = 6: 4: 3`
The lenght of wire L = 260 cm , so
`l_(1) =(6)/(13) xx 260 = 120 cm `
` l_(2) = (4)/(13) xx 260 = 80cm`
and `l_(3) =(3)/(13) xx 260= 60 cm`