A) \[n={{n}_{1}}+{{n}_{2}}+{{n}_{3}}+...\]
B) \[{{n}^{2}}=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}+...\]
C) \[\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}+...\]
D) \[\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{{{n}_{1}}}}+\frac{1}{\sqrt{{{n}_{2}}}}+\frac{1}{\sqrt{{{n}_{3}}}}+...\]
Correct Answer: C
Solution :
From law of length, the frequency of vibrating string is inversely proportional to its length, i.e., \[n\propto \frac{1}{l}\] or \[nl=\] constant (say k) or \[nl=k\] or \[l=\frac{k}{n}\] The segment of string of length \[{{l}_{1}},\,{{l}_{2}},\,{{l}_{3}}...\] have frequencies \[{{n}_{1}},{{n}_{2}},{{n}_{3}}\,,....\] Total length of string is \[l\]. So, \[l={{l}_{1}}+{{l}_{2}}+{{l}_{3}}+...\] \[\therefore \frac{k}{n}=\frac{k}{{{n}_{1}}}+\frac{k}{{{n}_{2}}}+\frac{k}{{{n}_{3}}}+.....\] \[or\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}+....\]You need to login to perform this action.
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