A uniform rod of length `L` (in between the supports) and mass `m` is placed on two support `A` and `B`. The rod breaks suddenly at length `L//10` from the support `B`. Find the reaction at support `A` immediately after the rod breaks

Options

A. `9/40 mg`
B. `19/40 mg`
C. `(mg)/2`
D. `9/20 mg`

Answer

A

Explanation

Torque `=tau=9/10mg(9/20L)`
`=Ialpha=m/3(9/10L)^(2) alpha,a=(3g)/(2L)`
Acceleration
`a_(CM)=alpha(AC)=(3g)/(2L)((9L)/20)=(27g)/40`
`N_(A)=9/40mg`