Just after cutting the string the tension of right will disapear and tension in left string will change. Let us draw free body diagram of rod immediately after the string breaks.
`SigmaF_(x)=0`
`:. a_(x)=0`…….i
`a_(y)=(Sigma F_(y))/m=(mg-T)/m`........ii
Applying torque equation about centre of mass
`tau=I_(cm)alpha`
`T.(Lcos30^@)=(3(2L)^(2))/12.alpha`
`implies alpha=tau/Ii=(TLcos30^@)/((m(2L)^(2))/12)=(3sqrt(3)T)/(2mL)` ..........iii
Now just after the string breaks, acceleration of point `A` in the vertical direction should be zero.
`i.e. a_(y)=Lalpha cos30^@ `.........iv
Solving equatioin i, ii, iii and iv we get
`T=(2mg)/11` and `a=(3sqrt(3))/11 g/L`