A) \[\frac{n}{4}\]
B) \[\sqrt{2}\,n\]
C) \[n\]
D) \[\frac{n}{\sqrt{2}}\]
Correct Answer: C
Solution :
\[n=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}\rho }}\propto \sqrt{\frac{T}{{{r}^{2}}\rho }}\] \[\Rightarrow \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right){{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\left( \frac{{{\rho }_{2}}}{{{\rho }_{1}}} \right)}\] \[=\sqrt{\left( \frac{1}{2} \right){{\left( \frac{2}{1} \right)}^{2}}\left( \frac{1}{2} \right)}=1\] \[\therefore \] \[{{n}_{1}}={{n}_{2}}\]
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