Properties of Matter

$ \begin{aligned} &\text { Equation the forces, we get }\\ &\begin{gathered} \mathrm{P}_0 \pi\left(4 r^2\right)+(\rho / 3) 4 \pi r^2 h g=\left[\mathrm{P}_0+\rho g h_2\right] \pi\left(4 r^2\right) \\ (\rho / 3) 4 \pi r^2 h g=\rho g h_2 3 \pi r^2 \end{gathered} \end{aligned} $

$ h_2=\frac{4 h}{9} $

For $h_2<\frac{4 h}{9}$, Cylinder does not move up because further buoyant force decreases while the weight of block remains same.

Weight of liquid column of height $\mathrm{H}, \mathrm{W}=\rho \mathrm{V} g$

$=\frac{\rho\left(\pi d^{2}\right)}{4} \mathrm{Hg}$ ....... (i)

Let us consider a small element $d x$ of mass $d m$ at a distance $x$ (as shown in figure).

Centripetal force required, in rotating this mass $=d m \omega^{2}$

$\therefore$ Total force required to rotate mass of length $\mathrm{L}$,

$\begin{aligned} \mathrm{F} & =\int d m \omega^{2} \\ & =\int_{0}^{\mathrm{L}} \frac{\rho\left(\pi d^{2}\right)}{4} x d x \omega^{2} \\ & =\frac{\rho\left(\pi d^{2}\right)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2} ~~........ (\mathrm{ii})\end{aligned}$

This centripetal force is provided by weight of liquid column (of height H), Here on comparing equation (i) and (ii), we get

$\begin{aligned} W & =\mathrm{F} \\ \frac{\rho\left(\pi d^{2}\right)}{4} \mathrm{Hg} & =\frac{\rho\left(\pi d^{2}\right)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2} \\ \mathrm{H} & =\frac{\mathrm{L}^{2} \omega^{2}}{2 g} \end{aligned}$