question_answer

A string of length L is fixed at one end and carries a mass M at the other end.  The string makes 2/p revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is [BHU 2002; DPMT 2004]

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)             \[T=M{{\omega }^{2}}L\]             \[=M\,4{{\pi }^{2}}{{n}^{2}}L\]             \[=M\,4{{\pi }^{2}}{{\left( \frac{2}{\pi } \right)}^{2}}L\]             \[=16\,ML\]