AITS phaltu question.

this is an easy one(but in the exam due to different data in question and in figure i didnt take the risk)
first of all the velocity along the rod's length is zero(as its length remains constant)

hence let the rod be r^->

and both the velocities be a^-> and b^->

then just apply
r^->.a^-> and r^->.b^->

and from there find the velocities along the rod (which is r^->.a^->/2 and r^->.b^->/2)

equating them

2√3+Vy=3√3+6
Vy=6-√3

now find the perpendicular velocities at both the ends

and then find the instantaneous centre of rotation and apply wx=Va or Vb

Most probably they say in FIITJEE... this question was cancelled...

As far as this question is concerned The length of rod cannot increase thats the trick...

can u please tell me wats phaltu!!!!!!

i dont know hindi

can anyone expain to me celestines working

i think its some kind of shortcut

this question was wrong and was cancelled last year.....

lolz..rofl