A) \[\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}\]
B) \[\frac{1}{\sqrt{n}}=\frac{1}{{{\sqrt{n}}_{1}}}+\frac{1}{\sqrt{{{n}_{2}}}}+\frac{1}{\sqrt{{{n}_{3}}}}\]
C) \[\sqrt{n}=\sqrt{{{n}_{1}}}+\sqrt{{{n}_{2}}}+\sqrt{{{n}_{3}}}\]
D) \[n={{n}_{1}}+{{n}_{2}}+{{n}_{3}}\]
Correct Answer: A
Solution :
[a] \[{{n}_{1}}=\frac{v}{2{{\ell }_{1}}}\Rightarrow {{\ell }_{1}}=\frac{v}{2{{n}_{1}}}.\] Similarly \[{{\ell }_{2}}=\frac{v}{2{{n}_{2}}},{{\ell }_{3}}=\frac{v}{2{{n}_{3}}}\] For complete wire \[\ell =\frac{2v}{2n}\] \[\ell ={{\ell }_{1}}+{{\ell }_{2}}+{{\ell }_{3}}\] \[\frac{v}{2n}=\frac{v}{2{{n}_{1}}}+\frac{v}{2{{n}_{2}}}+\frac{v}{2{{n}_{3}}}\] \[\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}\]
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