A) \[\sqrt{v}=\sqrt{{{v}_{1}}}+\sqrt{{{v}_{2}}}+\sqrt{{{v}_{3}}}\]
B) \[v={{v}_{1}}+{{v}_{2}}+{{v}_{3}}\]
C) \[\frac{1}{v}=\frac{1}{{{v}_{1}}}+\frac{1}{{{v}_{2}}}+\frac{1}{{{v}_{3}}}\]
D) \[\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{{{v}_{1}}}}+\frac{1}{\sqrt{{{v}_{2}}}}+\frac{1}{\sqrt{{{v}_{3}}}}\]
Correct Answer: C
Solution :
The fundamental frequency of string \[v=\frac{1}{2l}\sqrt{\frac{T}{m}}\] \[\therefore \] \[{{v}_{1}}{{l}_{1}}={{v}_{2}}{{l}_{2}}={{v}_{2}}{{l}_{3}}=k\] ?(i) From Eq. (i) \[{{l}_{1}}=\frac{k}{{{v}_{1}}},{{l}_{2}}=\frac{k}{{{v}_{2}}},{{l}_{3}}=\frac{k}{{{v}_{3}}}\] Original length\[l=\frac{k}{v}\] Here, \[\begin{align} & l={{l}_{1}}+{{l}_{2}}+{{l}_{3}} \\ & \frac{k}{v}=\frac{k}{{{v}_{1}}}+\frac{k}{{{v}_{2}}}+\frac{k}{{{v}_{3}}} \\ & \frac{1}{v}=\frac{1}{{{v}_{1}}}+\frac{1}{{{v}_{2}}}+\frac{1}{{{v}_{3}}} \\ \end{align}\]
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