question_answer

A string of length \[L\] is fixed at one end and carries a mass \[M\] at the other end. The string makes \[\frac{2}{\pi }\] revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

A) \[ML\]                                 

B) \[2ML\]

C) \[4ML\]                               

D) \[16ML\]

Correct Answer: D

Solution :

Resolving forces                 \[T\sin \theta =m{{\omega }^{2}}R\] or            \[T\sin \theta =m{{\omega }^{2}}L\sin \theta \] \[\therefore \]  \[T=m{{\omega }^{2}}L=m4{{\pi }^{2}}{{n}^{2}}L\]                 \[=m4{{\pi }^{2}}{{\left( \frac{2}{\pi } \right)}^{2}}L=16ML\]