A) 1 : 4
B) 4 : 1
C) 2 : 1
D) 1 : 2
Correct Answer: C
Solution :
The fundamental frequency of a wire is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] where\[l\]is length of wire, T the tension and m the mass per unit length. \[m=\frac{mass\text{ }of\text{ }wire}{length\text{ }of\text{ }wire}\] \[=\frac{\pi {{r}^{2}}L\times density}{L}=\pi {{r}^{2}}d\] \[n=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}d}}\] \[\Rightarrow \] \[n=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}\] \[\therefore \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{2}{1}\]
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