Due to rotation about the axis $O O^{\prime}$, a force is acting on the liquid column $A B$ outwards, pushing up the liquid column on the right. This centrifugal force acting on the small element $d x$ of area of cross-section $d x$, distant $x$ from the axis is
$$
d m \omega^2 x=A d x \rho \omega^2 x
$$
The total force due to column of length $L(=A B)$ is
$$
\int_0^L A \rho \omega^2 \cdot x d x=A \rho \omega^2 \frac{L^2}{2}
$$




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
$$
\begin{aligned}
& \quad \text { atm. pressure }+h_1 \rho g+\rho \omega^2 \frac{L^2}{2}=\text { atm. pressure }+h_2 \rho g \\
& \therefore \quad h_2-h_1=H_0=\frac{\omega^2 L^2}{2 g} .
\end{aligned}
$$

$$
\text { Here one has }
$$


$$
H_2=\frac{\omega^2 L_2^2}{2 g}, \quad H_1=\frac{\omega^2 L_1^2}{2 g}
$$
As $\omega$ is the same, and $L_2>L_1, H_2>H_1$. But both will go up.