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Answer :

`(sqrt(2)-1)`Solution :

Moment of inertia of block about its edge `D` is equal to `2/3 Ma^(2)` <br> When bullet strikes the blocks and gets embedded into it even then moment of inertia remains practically equla to `I=2Ma^(2)//3` because `m lt lt M` <br> During the collision, a reaction is offered by `D`. Hence momentum of the system dies not remain constant but the angular momentum of the sytem remains constant about `D` because angular impulse produced by the reaction is zero about `D` itself. <br> If angular velocity of the block, just after the collision is `w` then according to law of coservation of angular momentum, <br> `I.omega=mv.a/2implies omega=(3mv)/(4Ma)` <br> The block will jus topple, if the block reaches to the positioin shown in figure <br> Here `/_\PE=-/_\KE` of the system <br> `Mg(a/sqrt(2)-a/2)=1/2Iomega^(2)` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C03_E01_077_S01.png" width="80%"> <br> `implies v=(2M)/Msqrt(2/3 ga(sqrt(2)-1))`