Correct Answer - A
Let the length of a small element of tube be `dx` . Mass of this element
`dm=(M)/(L)dx`
Where `M` is mass of filled liquid and `L` is the length of tube.
Force on this element
`dF=(dm)omega^(2)x=((M)/(L))dx.omega^(2)x`
Integrating
`"in"t_(0)^(F)dF=(M)/(L)omega^(2)"in"t_(0)^(L)xdx`
or `F=(M)/(L)omega^(2)[(L^(2))/(2)]=(MLomega^(2))/(2)`
or `F=(1)/(2)MLomega^(2)` .