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 immediately after string breaks is

Answer

4

Explanation

`sumF_(x)=0,a_(y)(sum F_(Y))/(m)=(mg-T)/m`
`alpha=tau/I=(TL cos30^(@))/((m(2L)^(2))/12)=(3sqrt(3)T)/(2ml)`
Now just after the string breaks acceleration of point `A` in vertical direction should be zero solving above equtions we get
`T=4/13 mg` and `alpha=(6sqrt(3))/(13L)`