Correct Answer - A
The force acting on the mass of liquid of length `dx` at a distance `x` from the axis of rotation `O` is as follows:
`dF=(dm)xxomega^(2)`
`:. dF=M/Ldx xxomega^(2)`. Therefore the force action at the other end is for the whole liquid in tube.
`Fint_(0)^(L)M/L omega^(2) x dx=M/L omega^(2) int_(0)^(L) xdx`
`=M/Lomega^(2)[(x^(2))/2]^(L)=M/Lomega^2[(L^(2))/2-0]`
`=(MLomega^(2))/2`