A rod of length ‘$l$’ is travelling with velocity ‘v’ and rotating with angular velocity ‘$\omega$’ such that $\omega$$l$/2=v. The distance travelled by end B of the rod when the rod rotated by an angle,   is (x$l$) .Find the value of x is

Here the point ‘B’ is following a curvilinear path whose velocity at any instant is
given by vector sum of $\overrightarrow{v}$ and   When the rod rotates by an angle   the
situation is as shown in the figure. The speed of end point at this instant is
 

$=\sqrt{2}v\left(\sqrt{1+\cos \theta }\right)$    
 
 
${\int }_0^{s}ds=l{\int }_0^{\pi /2}\cos \left(\theta /2\right)d\theta =\frac{2l}{\sqrt{2}}=\sqrt{2}l$