A solid paraboloid of base radius R0 and height h0 is having uniform volume mass density of ρs, is inverted and placed just above the surface as shown, on a liquid having volume mass density of ρL. When paraboloid is released from rest it has been observed that when it becomes completely submerged in the liquid the paraboloid comes to the rest again. There exist uniform gravity of g. Assume that the liquid body is very large so that liquid level is not change as the paraboloid enters into the liquid. The value of ratio ρLρs is

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A solid paraboloid of base radius $R_{0}$ and height $h_{0}$ is having uniform volume mass density of $ρ_{s}$ , is inverted and placed just above the surface as shown, on a liquid having volume mass density of $ρ_{L}$ . When paraboloid is released from rest it has been observed that when it becomes completely submerged in the liquid the paraboloid comes to the rest again. There exist uniform gravity of g. Assume that the liquid body is very large so that liquid level is not change as the paraboloid enters into the liquid. The value of ratio $\frac{ρ_{L}}{ρ_{s}}$ is

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answer is 3.

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Detailed Solution

Let $Y = a x^{2}$ (A parabola)
$Y = \frac{H}{R^{2}} x^{2}$ (from given data)

$v = \frac{d y}{d t} m g - ρ_{L} . \frac{1}{2} π x^{2} y . g = m . \frac{d v}{d t} \frac{d v}{d t} = g - \frac{π x^{2} y g ρ_{L}}{2 ρ_{S} \frac{π R^{2} H_{o}}{2}} g \Rightarrow g - \frac{ρ_{L}}{ρ_{S}} \frac{x^{2}}{R^{2}} \times \frac{y g}{H_{o}} v . \frac{d v}{d t} = g - \frac{ρ_{L}}{ρ_{S}} \frac{g y^{2}}{{H_{o}}^{2}} \int_{0}^{0} v d v = \int_{0}^{H_{o}} (g - \frac{ρ_{L}}{ρ_{S}} g y^{2}) d y 0 = g H_{o} - \frac{g ρ_{L}}{3 ρ_{S}} H_{o} \frac{ρ_{L}}{ρ_{S}} = 3$