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 string is :

A)  \[16\,ml\]

B)  \[4\,ml\]

C)  \[8\,ml\]

D)  \[2\,ml\]

Correct Answer: A

Solution :

 Key Idea: Horizontal component of tension balances centripetal force. The free body diagram of the given situation is shown. Taking the vertical and horizontal component of forces, we have \[T\sin \theta =\frac{m{{v}^{2}}}{r}\] ... (i) \[T\cos \theta =mg\] ... (ii) where linear velocity \[v=r\,\,\omega \] and \[\sin \theta =\frac{r}{l}\] Putting these values in (i), we get \[T\times \frac{r}{l}=m{{\omega }^{2}}T\] We know \[\omega =2\,\pi \,n\], we have \[\therefore \] \[T=m\,{{(2\pi n)}^{2}}\,l\] \[\Rightarrow \] \[T=m\,{{\left( 2\pi \times \frac{2}{\pi } \right)}^{2}}l\] \[\Rightarrow \] \[T=16\,mL\].