A) \[\frac{3}{8}\]
B) \[\frac{2}{3}\]
C) \[\frac{8}{9}\]
D) \[\frac{9}{4}\]
Correct Answer: D
Solution :
\[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] \[\Rightarrow \] \[n\propto \frac{\sqrt{T}}{l}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left[ \frac{{{n}_{2}}}{{{n}_{1}}} \right]}^{2}}{{\left[ \frac{{{l}_{2}}}{{{l}_{1}}} \right]}^{2}}\] \[={{(2)}^{2}}\left[ \frac{3}{4} \right]=\frac{9}{4}\]You need to login to perform this action.
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