A) 8 times
B) 4 times
C) 1/3 times
D) half times
Correct Answer: B
Solution :
Here, initial vibration \[{{n}_{1}}=n\] Final vibration \[{{n}_{2}}=2n\] Initial tension \[T=T\] The vibration of frequency of string is \[n=\frac{1}{2l}\frac{\sqrt{T}}{m}\propto \sqrt{T}\] Hence, \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\sqrt{{{T}_{1}}}}{{{T}_{2}}}\] or \[\frac{n}{2n}\frac{\sqrt{{{T}_{1}}}}{{{T}_{2}}}\] or \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{4}\] or \[{{T}_{2}}=4{{T}_{1}}\]
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