Correct Answer - A
`AC=asqrt(2)`
The plates falls by `AO=a//sqrt(2)`
`I`(diagonal `DB`)=`(I_(z))/2=(I_(1)+I_(2))/2`
`=1/2[(Ma^(2))/2]=(Ma^(2))/6`
Since `I_(1)=I_(2)=(Ma^(2))/6,I_(A)=I_(DB)+M(AO)^(2)`
`=(Ma^(2))/6+M(a//sqrt(2))^(2)=(2/3)Ma^(2)=2/3Ma^(2)`
`Mga/(sqrt(2))=1/2Iomega^(2)=1/2(2/3)Ma^(2)omega^(2)`
Which gives `omega=[(3g)/(2a)]^(1//2)=[(3a)/(asqrt(2))]^(1//2)`