Correct Answer - 1
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the sam amount, `x`. If the density of the liquid is `rho`, then, the total mechanical energy of the liquid
coloum is: `E =(1)/(2) {A(hxx x) rho +A(h-x)rho}. ((dx)/(dt))^(2)`
`+[A(h+x) rho.g.(h+x)/(2)+A(h-x).rho.g.(h-x)/(2)]`
`=(1)/(2) (2Ahrho) ((dx)/(dt))^(2) +(1)/(2) (2A rho g) (h^(2)+x^(2))`
After differentiating the total energy and equating it to zero, one finds acceleration `=- omega^(2)x`.
The angular frequency of small oscillations, `omega` is:
`omega = sqrt((2A rhog)/(2A h rho)) = sqrt((g)/(h)) = sqrt((10)/(10)) =1 rad//s`