A) eight times
B) four times
C) twice
D) half
Correct Answer: B
Solution :
Here: initial vibration\[{{n}_{1}}=n\], final vibration\[{{n}_{2}}=2n\] Initial tension\[{{T}_{1}}=T\] Vibration of frequency of string \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] \[\Rightarrow \] \[n\propto \sqrt{T}\] Hence, \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] or \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{2}}\] \[={{\left( \frac{n}{2n} \right)}^{2}}=\frac{1}{2}\] or \[{{T}_{2}}=4{{T}_{1}}\]
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