Physics - Fluid Question with Solution | TestHub

Due to rotation about the axis $O{O}^{\prime }$, a force is acting on the liquid column $AB$ outwards, pushing up the liquid column on the right. This centrifugal force acting on the small element $dx$ of area of cross-section $dx$, distant $x$ from the axis is $$dm{\omega }^{2}x=Adx\rho {\omega }^{2}x$$ The total force due to column of length $L(=AB)$ is $${\int }_0^{L}A\rho {\omega }^2\cdot xdx=A\rho {\omega }^2\frac{L^2}{2}$$
Pressure at $B=$ atmospheric pressure + $$h_1\rho g+\frac{A\rho {\omega }^2}{A}\frac{L^2}{2}$$ Pressure at $B$ due to liquid column on the right hand side $=$ atmospheric pressure $+h_2\rho g$ Net pressure due to the left hand side $=$ pressure at $B$ due to liquid column $$\begin{array}{rl}& \text{ atm. pressure }+h_1\rho g+\rho {\omega }^2\frac{L^2}{2}=\text{ atm. pressure }+h_2\rho g\\ & \therefore h_2-h_1=H_0=\frac{{\omega }^2{L}^2}{2g}.\end{array}$$ $$\text{ Here one has }$$
$$H_2=\frac{{\omega }^2{L}_2^2}{2g},H_1=\frac{{\omega }^2{L}_1^2}{2g}$$ As $\omega$ is the same, and $L_2>L_1,H_2>H_1$. But both will go up.