Correct Answer - A
Let the system starts with agular velocity `omega`. Angular velocity if the ball will also be `omega` as it remais struck to the rod.
Velocity of the ball `v=omegaL/3`
For the rod angular impulse= change in angular momentum:
`J(2L)/3-J_(b) L/3=(ML^(2))/3omega`.....i
For the ball, impulse `=` change in linear momentum
`J_(b)=mv=momega(L/3)`......ii
From eqn i and ii `omega=(6J)/((m+3M)L)`
and `J_(b)=(2mJ)/((m+3M))`
Let impulse on the pivot be `J_(1)`
For the rod and ball system total impulse `=` change in linear
mometum `J+J=Mv_(c)+mv`
Solve to get `J_(1)=(mJ)/(m+3M)`