A spherical ball of radius R is made by joining two hemispherical parts. The two parts have density $ρ$ and $2 ρ$ . When placed in a water tank, the ball floats while remaining completely submerged. Neglect viscous forces. Mark the correct statement(s) :-

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Density of water is $ρ_{0}$ , then $ρ = \frac{2 ρ_{0}}{3}$

Position of centre of mass of ball is at distance of $\frac{R}{8}$ from O.

Moment of inertia of ball about COM of ball perpendicular to the plane of paper is $\frac{123}{160} {ρπR}^{5}$

Time period of small angular oscillations of the ball about its equilibrium position is given by $2 π \sqrt{\frac{123 R}{40 g}}$

answer is A, B, C, D.

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Detailed Solution

Where $M = f (\frac{2}{3} {πR}^{3})$

$\begin{array}{l} y_{cm} = \frac{M (3 R / 8) + 2 M (- 3 R / 8)}{3 M} \\ y_{cm} = - \frac{R}{8} \end{array}$

For equilibrium, F_B= weight of ball

$\begin{array}{l} ρ (\frac{4}{3} {πR}^{3}) = ρ (\frac{2}{3} {πR}^{3}) + 2 ρ (\frac{2}{3} {πR}^{3}) \\ 4 ρ_{0} = 6 p \\ ρ = \frac{2 ρ_{0}}{3} \\ I_{0} = \frac{2 M}{5} R^{2} + \frac{2 (2 M)}{5} R^{2} where M = \frac{2}{3} {πR}^{3} ρ \\ I_{0} = \frac{6 {MR}^{2}}{5} \\ I_{cm} = I_{0} - {md}_{2} \\ = \frac{6 {MR}^{2}}{5} - \frac{3 {MR}^{2}}{64} = \frac{369}{320} {MR}^{2} \\ = \frac{369}{320} R^{2} (\frac{2}{3} {πR}^{3} ρ) \\ I_{cm} = \frac{123}{160} {ρπR}^{5} \end{array}$