question_answer

A string of length\[l\]fixed at one end carries a mass m at the other end. The string makes \[\frac{2}{\pi }rev/s\]around the horizontal axis through the fixed end as shown in the figure, the tension in the string is

A) \[16\,ml\]            

B) \[6\,ml\]

C) \[5\,ml\]           

D)  \[3\,ml\]

Correct Answer: A

Solution :

 \[T\sin \theta =\frac{m{{v}^{2}}}{r}\] \[T\cos \theta =mg\] \[v=r\omega \]and \[\sin \theta =\frac{r}{l}\] Putting these values in Eq (i), we get \[T\times \frac{r}{l}=m{{\omega }^{2}}r\] \[T=m{{\omega }^{2}}l\] \[\omega =2\pi n\] \[T=m{{(2\pi n)}^{2}}l\] \[T=m{{\left( 2\pi \times \frac{2}{\pi } \right)}^{2}}l\] \[T=16\,ml\]