The correct option is C 32√gL
Moment of inertia of the system about the given axis,
I=IA+IB+IC ...(i)
The moment of inertia of rod A about axis passing through A will be zero, since the ⊥ distance of particles of rod from the axis of rotation will be zero (considering rod is thin).
∴IA=0
For the axis of rotation passing through rod A, rod B is rotating about its one end.
⇒IB=ML23
For rod C all the points on rod are at a ⊥ distance L from the axis of rotation, so
⇒IC=∑miL2
⇒IC=L2∑mi=ML2
Where mi is mass of each particle.
Substituting the values in Eq.(i),
⇒I=0+ML23+ML2=43ML2
let ω be the angular speed of body(H) when it is rotated from horizontal to vertical position.
Gain in rotational kinetic energy of body is,
ΔK.ERot=12Iω2=12[43ML2]ω2
⇒When body gets oriented to vertical position, the loss in gravitational P.E of individual rods is,
ΔP.EA=0, because position of centre of mass of rod remains same.
ΔP.EB=MgL2, because centre of mass of rod B gets lowered through a distance L2
ΔP.EC=MgL, centre of mass of rod C gets lowered through a distance L
⇒Loss in potential energy of the system during this process of rotation will be,
ΔP.Esystem=0+MgL2+MgL
⇒ΔP.Esystem=32MgL
Applying conservation of mechanical energy on the body:
(Gain in Rotational K.E equals the loss in gravitational P.E of system)
⇒ΔK.ERot=ΔP.Esystem
⇒23ML2ω2=32MgL
∴ω=32√gL