Let the initial angular velocity of the laminar sheet be w

Let w’ be the angular velocity with which sheet turns back after the impact with the hinge 

Now, Change in angular moment = moment of impulse

=> I_AB w - I_AB (-w’) = 0.5 * 6     

=> I_AB (w+w’) = 3 Nm-s …..............eq 1

Now, as per question,  the hinge doesn’t impart any impulse, so we can say that

     Change in linear momentum of laminar sheets = impulse on the sheet due to each obstacle

 => ml (w+w’) = 6 N-s ……..............eq 2

where l is the distance of the centre of mass of the laminar sheet from AB

   By using the theorem of parallel axes, we can find out the moment of inertia of the laminar sheet about AB 

     I_AB = I₀ +ml²                 ……........eq 3

Dividing eq 1 by eq 2 , we get

I_AB /ml =  1/2

=> (I₀ +ml² )/ml =1/2 (by using eq 3 )

=> 2ml² –ml +2I_o =0

put the value of m and Io from the question and solve the quadratic equation

Solving above quadratic equation , we get l =  0.4 m or 0.1 m

now, if we put l=0.4 in eq 2, we get negative w’ which is not possible (as the sheet must come back)

so we get l = 0.1m

corresponding w’ = 1 rad/s

ans. a) l=0.1 m

ans. b) w’= 1 rad/s

ans. c) Sheet will never come to rest