GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

For equilibrium
Buoyant force = weight of hollow sphere

${\mathit{\rho }}_{\mathit{\omega }}\times \frac{\mathit{V}}{2}\times \mathit{g}={\mathit{\rho }}_5\times \mathit{V}\mathit{g}$
${\mathit{\rho }}_{\mathit{s}}=\frac{{\mathit{\rho }}_{\mathit{\omega }}}{2}$
Mass of hollow sphere $={\mathit{\rho }}_{\mathit{s}}\times \mathit{V}$
$=\frac{{\mathit{\rho }}_{\mathit{\omega }}}{2}\times \mathit{V}$
$=500\times \frac{4}{3}\mathit{\pi }\times {\left(0.2\right)}^3$
$=16.7551\mathit{k}\mathit{g}$
Area cut by water surface in area of circle $={\mathit{n}\mathit{R}}^2$
Now the sphere is displaced by small $\mathit{x}$ so net disturbing force acting on the spherical ball for small oscillation $={\mathit{\rho }}_{\mathit{\omega }}\mathit{\pi }{\mathit{R}}^2\mathit{g}\mathit{x}$
Inertia force $=\mathit{m}¨\mathit{x}$
By D-Alembert's principle:
${\mathit{\omega }}_{\mathit{n}}=\sqrt{\frac{{\mathit{\rho }}_{\mathit{\omega }}\mathit{\pi }{\mathit{R}}^2\mathit{g}}{\mathit{m}}}$
$=\sqrt{\frac{{10}^3\times \mathit{\pi }\times {\left(0.2\right)}^2\times 10}{16.7551}}$
$=\sqrt{75}=8.66\mathit{r}\mathit{a}\mathit{d}/\mathit{s}\mathit{e}\mathit{c}$