A solid paraboloid of base Radius R and height h, having uniform volume mass density $ρ_{1}$ is inverted and just placed above the surface of a liquid of uniform volume mass density $ρ_{2}$ as shown in fig. When left free, its maximum displacement is equal to h and it gets submerged just completely. Assume that liquid body is large enough, that level of liquid doesn’t change when paraboloid enters the liquid (Neglect viscosity and if any dissipative forces of liquid). Then which of the following are correct?

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$\frac{ρ_{2}}{ρ_{1}} = 3$

It oscillates between extremes simple harmonically.

It oscillates between these extremes, but Oscillations are not SHM.

Submerged depth at equilibrium = $\frac{h}{\sqrt{3}}$ , Maximum speed in the oscillation = $V = \sqrt{\frac{4 g h}{3 \sqrt{3}}}$

answer is A, C, D.

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Detailed Solution

From fig; $y = \frac{h}{R^{2}} x^{2}$ ; $v o l u m e = \frac{1}{2} π R^{2} h$
At any instant, Let y be the submerged depth, then equation of motion $\Rightarrow m \frac{d v}{d t} = m g - (\frac{1}{2} π x^{2} y) ρ_{l} g$

$v \frac{d v}{d y} = g - \frac{ρ_{l}}{ρ_{s}} \frac{g}{h^{2}} y^{2}$

$y_{\max} = h \Rightarrow g h = \frac{ρ_{l}}{ρ_{s}} \frac{g}{h^{2}} \frac{h^{3}}{3} \Rightarrow \frac{ρ_{l}}{ρ_{s}} = 3$ ………………..(A)
At equilibrium, a=0

$m g = \frac{1}{2} π x^{2} y_{0} . ρ_{l} g$

$\frac{1}{2} π R^{2} h ρ_{s} g = \frac{1}{2} π x^{2} y_{0} . ρ_{l} g$

$\Rightarrow {y_{0}}^{2} = h^{2} \frac{ρ_{s}}{ρ_{l}} \Rightarrow y_{0} = \frac{h}{\sqrt{3}}$ …………………(D)
$\frac{v^{2}}{2} = g y - \frac{3 g}{h^{2}} \frac{y^{3}}{3}$