A) 2 : 1
B) 4 : 1
C) 3 : 2
D) 8 : 1
E) 4 : 5
Correct Answer: D
Solution :
\[n=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}d}}\]where\[l\]is length, T is tension, r is radius and d is density. Given, \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{2},\]\[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{1}{4}\] \[\therefore \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\sqrt{\frac{r_{2}^{2}}{r_{1}^{2}}}\] \[\Rightarrow \] \[{{\left( \frac{{{n}_{1}}{{l}_{1}}}{{{n}_{2}}{{l}_{2}}} \right)}^{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{n}_{2}}{{l}_{2}}}{{{n}_{1}}{{l}_{1}}}=2\times 4=8:1\]
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