Figure shows a right angle rod (having one arm vertical and other horizontal ) . The acceleration of free end A just after the system is released in `(xsqrt(2)g)/10` then x will be

Answer

3

Explanation

`tau=(mg)/2 l/2`
`=(m/2xx(l^(2))/3+m/2 (l^(2))/12 +m/2 xx(l^(2)+(l^(2))/4))alpha`
`(mgl)/4=5/6 m12 alpha`
`alpha=(3g)/(10 l)`
`a_(A)=sqrt(2)l alpha =(3sqrt(2)g)/10`