A string of length 1m is fixed at one end with a bob of mass 100 g and the string makes $\left(\frac{2}{\pi }\right)\text{ rev }{s}^{-1}$ around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

$T\cos \theta =mg$                                                    …(i)
$T\sin \theta =mr{\omega }^2=m\left(l\sin \theta \right){\omega }^2$                      …(ii)
Solving these two equations we get,
 $\cos \theta =\frac{g}{l{\omega }^2}=\frac{g}{l{\left(2\pi n\right)}^2}$
 $=\frac{10}{{\left[2\pi \text{x}\frac{2}{\pi }\right]}^2}(l=1\text{ m})$
 $\therefore \theta ={\cos }^{-1}(5/8)$