In the arrangement shown, the end B of the rod of length  ‘l’ is made to move with constant velocity V. Then velocity of its centre of mass when θ=300 is

O is the position of instantaneous centre of rotation. Then entire rod is rotating about O with angular velocity ω at that instantwhere ω = vBOB  =  vl sin θNow    OA2  +  OB2 = 2.  AC2  +  OC2⇒     l2 cos2θ  +  l2  sin2 θ  =  2 l24  +  OC2⇒   OC2  = l22  −  l24 =  l24      ⇒   OC = l2∴     vC =  ω . OC  =  vl  sin  θ  ×  l2  =  v2 sin θWhen θ  = 300,   vC  =  v2 sin 300 =  v