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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 beA. 80 cm, 60 cm, 120 cmB. 120 cm, 80 cm, 60 cmC. 60 cm, 80 cm, 120 cmD. 120 cm, 60 cm, 80 cm |
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Answer» Correct Answer - B 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` |
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