A) ML
B) 2 ML
C) 4 ML
D) 16 ML
Correct Answer: D
Solution :
\[T\sin \theta =M{{\omega }^{2}}R\] ?(i) \[T\sin \theta =M{{\omega }^{2}}L\sin \theta \] ?(ii) From (i) and (ii) \[T=M{{\omega }^{2}}L\] \[=M\,4{{\pi }^{2}}{{n}^{2}}L\] \[=M\,4{{\pi }^{2}}{{\left( \frac{2}{\pi } \right)}^{2}}L\] \[=16\,ML\]You need to login to perform this action.
You will be redirected in
3 sec