A uniform rod of mass `M = 2 kg` and length `L` is suspended by two smooth hinges `1` and `2` as shown in Fig. A force `F = 4 N` is applied downward at a distance `L//4` from hinge `2`. Due to the application of force `F`, hinge `2` breaks. At this instant, applied force `F` is also removed. The rod starts to rotate downward about hinge `1`. (`g = 10 m//s^(2)`)

The reaction at hinge `1`, just after breaking of hinge `2`, is

Options

A. `30m//s^(2)`
B. `20m//s^(2)`
C. `10m//s^(2)`
D. 0

Answer

A

Explanation

By conservation of energy
`Mg(L/2)=1/2Iomega^(2)=1/2((ML^(2))/3)omega^(2)`
`:.omega=sqrt((3g)/L)`
Acceleration `=omega^(2)L=3g=30m//s^(2)`