A thin uniform bar of mass `m` and length `2L` is held at angle `30^(@)` with the horizontal by means of two vertical inextensible strings, at each and as shown in figure. If the string at the right end breaks, leaving the bar to swing the tension in the string at the left end of the bar immediately after string breaks is `T=n/13 mg`

Explanation

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`