A manometer contains a liquid of density $\mathrm{\rho }$. The difference in the levels when the manometer rotates at a constant angular velocity $\mathrm{\omega }$ about one of its vertical limbs. (Let l be the length of horizontal tube)

Let

S = Lower limb cross-section

dx = the liquid element's length

dm = Mass of the liquid element = $\rho s dx$

$a_n=Acceleration = {\omega }^{2}x$

p to p + dp are the pressures at the end of the mass element. Range of distances: x to x + dx

$$\begin{gathered} \mathrm{Force}, \\ {\mathrm{F}}_{\mathrm{n}}=\mathrm{ma}; \end{gathered}$$

$(\mathrm{p}+\mathrm{dp})\mathrm{s} - \mathrm{ps} = \mathrm{\rho } (\mathrm{s} \mathrm{dx}) {\mathrm{\omega }}^2\mathrm{x}$

$\mathrm{At} \mathrm{point} \mathrm{O}: {\mathrm{P}}_{\mathrm{o}}=\mathrm{\rho } \mathrm{g} {\mathrm{y}}_1$

$\mathrm{At} \mathrm{point} \mathrm{B}: {\mathrm{P}}_{\mathrm{B}}=\mathrm{\rho } \mathrm{g} {\mathrm{y}}_2$

${\int }_{{\mathrm{P}}_0}^{\mathrm{P}}\mathrm{dp}={\int }_0^{\mathrm{L}}{\mathrm{\rho \omega }}^2\mathrm{xdx}$

$\mathrm{g}({\mathrm{y}}_2-{\mathrm{y}}_1)=\frac{{\mathrm{\omega }}^2{\mathrm{L}}^2}{2}$

$({\mathrm{y}}_2-{\mathrm{y}}_1)=\frac{{\mathrm{\omega }}^2{\mathrm{L}}^2}{2\mathrm{g}}$

Hence the correct answer is $\frac{{\mathrm{\omega }}^2{\mathrm{L}}^2}{2\mathrm{g}}.$