A) 20 N
B) 40 N
C) 10 N
D) 120 N
Correct Answer: B
Solution :
For mass m per unit length of wire and tension T, the frequency of note emitted by the wire is \[n=\frac{1}{2l}\,\sqrt{\frac{T}{m}}\] where l is length of wire. Given, \[{{\operatorname{T}}_{1}}=10N,\,\,{{n}_{1}}=n,\,\,and\,\,{{n}_{2}}=2n\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\Rightarrow \,\,\,\,\,\,\frac{n}{2n}=\sqrt{\frac{10}{{{T}_{2}}}}\] \[=\,\,{{T}_{2}}=10\,\,\times \,\,4\,\,-\,\,40\,N\]
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