A) 16 : 9 : 4 : 1
B) 4 : 3 : 2 : 1
C) 1 : 4 : 2 : 16
D) 1 : 2 : 3 : 4
Correct Answer: D
Solution :
Ratio of tensions of stretched wires \[{{T}_{1}}:{{T}_{2}}:{{T}_{3}}:{{T}_{4}}:\,=1:4:9:16\] The relation for fundamental frequency of stretched wire is given by \[{{v}_{1}}=\frac{1}{2l}\sqrt{\frac{T}{m}}\propto T\] Hence, ratio of fundamental frequencies of wires is given \[{{v}_{1}}:{{v}_{2}}:{{v}_{3}}:{{v}_{4}}:=\sqrt{{{T}_{1}}}:\sqrt{{{T}_{2}}}:\sqrt{{{T}_{3}}}:\sqrt{{{T}_{4}}}:\] \[=\sqrt{1}:\sqrt{2}:\sqrt{9}:\sqrt{16}\] \[=1:2:3:4\]You need to login to perform this action.
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