A) \[ML\]
B) \[2ML\]
C) \[4ML\]
D) \[16ML\]
Correct Answer: D
Solution :
Resolving forces \[T\sin \theta =m{{\omega }^{2}}R\] or \[T\sin \theta =m{{\omega }^{2}}L\sin \theta \] \[\therefore \] \[T=m{{\omega }^{2}}L=m4{{\pi }^{2}}{{n}^{2}}L\] \[=m4{{\pi }^{2}}{{\left( \frac{2}{\pi } \right)}^{2}}L=16ML\]You need to login to perform this action.
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