Diving Paraboloid

Let us call:

• $m$ as the paraboloid mass
• $R$ as the paraboloid base radius
• $H$ as the paraboloid height
• $y$ as the portion of the paraboloid height which is submerged
• $x$ as the radius of the base of the paraboloid sector which is submerged
• $v$ as the paraboloid vertical velocity downwards
• $\rho_{H2O}$ as the water density
• $\rho_{par}$ as the paraboloid density

First, if we assume that the paraboloid follows the relation by a constant $a$ :

$\large \displaystyle H = aR^2$

Then, also:

$\large \displaystyle y = ax^2$

Dividing both equations:

$\color{#20A900} \boxed{ \large \displaystyle x^2 = \frac{R^2}{H} y }$

Writing the differential equation of the paraboloid, since its displaced volume is $\frac{\pi x^2 y}{2}$ :

$\large \displaystyle m \frac{dv}{dt} = mg - \rho_{H2O} g \frac{\pi x^2 y}{2}$

$\large \displaystyle \frac{dv}{dt} = g - \rho_{H2O} g \frac{\pi R^2}{2mH} y^2$

Let us define:

$\color{#D61F06} \boxed{\large \displaystyle k \triangleq \rho_{H2O} g \frac{\pi R^2}{2mH} }$

$\large \displaystyle \frac{dv}{dt} = g - k y^2$

$\large \displaystyle \frac{dv}{dy} \frac{dy}{dt} = g - k y^2$

$\large \displaystyle \frac{dv}{dy} v = g - k y^2$

$\large \displaystyle v dv = \left( g - k y^2 \right) dy$

Since at $y=0$ , $v$ is also $0$ (paraboloid begins at rest):

$\large \displaystyle \int_0^v v dv = \int_0^y \left( g - k y^2 \right) dy$

$\color{#20A900} \boxed{ \large \displaystyle \frac{v^2}{2} = gy - \frac{ky^3}{3} }$

At $y = H$ (instant in which paraboloid is fully submerged), $v$ is also equal to $0$ :

$\large \displaystyle 0 = gH - \frac{kH^3}{3}$

$\large \displaystyle H^2= \frac{3g}{k}$

$\large \displaystyle H^2= 3g \frac{2mH}{\pi R^2 g \rho_{H2O} }$

$\large \displaystyle \frac{3m}{\frac{\pi R^2 H}{2}} = \rho_{H2O}$

$\large \displaystyle 3 \rho_{par} = \rho_{H2O}$

$\color{#3D99F6} \boxed{ \large \displaystyle \frac{\rho_{par}}{\rho_{H2O}} = \frac13 = 0.33}$

As it is not mentioned what sort of a paraboloid is taken ,i considered a simple paraboloid obtained by the rotation of parabola $y=x^2$ about the $y axis$ . Therefore the equation becomes $y=x^2+z^2$ . Now we know that the volume of a paraboloid with height $h$ and radius $a$ is $1/2πa^2h$ . Assume that $l$ length is inside the water already. Now we can find the volume submerged using the above mentioned results.

Next proceeding as your similar last question we obtain the asked ratio to be $1/3=0.33$