Uniform rod `AB` is hinged at the end `A` in a horizontal position as shown in the figure `(` the hinge is frictionless that is , it does not exert any friction force on the rod `)`. The other end of the rod is connected to a block through a massless string as shown. The pulley is smooth and massless . Masses of the block and the rod are same and are equal to `'m'`.

Then just after release of block from this position, the tension in the thread is

Options

A. `(mg)/(8)`
B. `(5mg)/(8)`
C. `(11mg)/(8)`
D. `(3mg)/(8)`

Answer

B

Explanation

Let `alpha` be the angular acceleration of rod and `a` be acceleration of block just after its release.
`:. Mg-T=ma …….(1)`
`Tl=mg``(l)/(2)=(ml^(2))/(3)alpha ....(2)`
and `a=l alpha ….(3)`

Solving we get
`T=(5mg)/(8) `and `alpha=(3g)/(8l)`
Now from free body diagram of rod, let `R` be the reaction by hinge on rod.
`R+T-mg=ma_(cm)=m(1)/(2) alpha`

Solving we get `R=(9mg)/(16)`