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

vel of end2 wrt 1 is 1i + 6-V j
now its perpendicular to rod

ωLsin30 = 1

ω=1

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.....

yeah i guess it was cancelled.....tats why i guess everyone got more marks then expected in PT1 last year.....those 12 marks were given as a gift[3]

lolz..rofl

I think the answer is 1 (a)

@kartik
see post no. 9
lolz