Q. 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
Jharkhand CECEJharkhand CECE 2014
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 $
