Uniform rod AB is hinged at end A in horizontal position as shown in the figure. The other end is connected to a block through a massless string as shown. The pulley is smooth and massless. Mass of block and rod is same and is equal to `m` Then acceleration of block just after release from this position is

Options

A. `(a)6g//13`
B. `(b)g//4`
C. `(c)3g//8`
D. (d)None of these

Explanation

`a=(mg-T)/(m)` .. (i)
`alpha=(tau)/(I)=(T.l-mg.(l)/(2))/((ml^(2)//3))`
or `alpha=(1)/(ml)[3T-(3mg)/(2)]` ..(ii)
`a=lalpha` …(iii)
solving these three equations we get,