A thin uniformed rod of mass `m` and length `L` is hinged at one end and from other end a light string is attached. The string is wound over a frictionless pullely (having mass `2m`) and a block of mass `2m` is connected to string on other side of pulley as shown. the system is released from rest when the rod id making an angle of `37^(@)` with horizontal. Based on above information answer the following question :

Just after releases of the system from rest acceleration of block is

Options

A. `(72g)/121,` downwards
B. `(48g)/119`, downwards
C. `(90g)/121`, downwards
D. `(90g)/121`,upwards

Answer

A

Explanation

Let `alpha` be the angular acceleration of rod about hinge and `a` be the acceleration of block just after the system is released from rest.
The free body diagram of rod, pulley and block are shown in figure. in free body diagram of rod, the forece exerted by hinge on rod is not shown. For rod, `mgxxL/2 cos 37^(@)-TxxL cos37^(@)`
`=-Ialpha[I=(mL^(2))/3]`
For block, `2mg-T=2ma`
From constraint, `a=Ialphacos37^(@)=(4Lalpha)/5`
Solving above equations, we get
`a=(72g)/121, alpha=(90g)/(121L),T=(98mg)/121`
Reactions force exerted by `H_(2)` on pulley
is, `N_(1)=2T+2mg=438/121mg`
Now draw the complete fee body diagram of rod as follows

`rArrR_(1)+Tcos37^(@)-mgcos37^(@)=(mL)/2 alpha`
`rArrR_(2)+T sin 37^(@)=mg sin 37^(@)`
Net reactions force on rod due to hinge is,
`F=sqrt(R_(1)^(2)+R_(2)^(2))`