Question

A composite rod whose upper half has a density $\frac{\rho }{4}$ and lower half has a density of $\frac{3\rho }{2}$ is immersed vertically in a liquid of density $\rho$. To what length, it should be immersed so that centre of buoyancy coincides with centre of mass of the rod? Suppose the length is $\frac{xL}{14}$. Find $x$.

Answer 1

Position of COM from bottom of the rod is
$Y_{cm}=\frac{\frac{L}{2}.A\times \frac{3\rho }{2}.\frac{L}{4}+\frac{L}{2}A.\frac{\rho }{4}.\frac{3L}{4}}{\frac{L}{2}A.\frac{3\rho }{2}+\frac{L}{2}A.\frac{\rho }{4}}=\frac{9L}{28}$
For COm to concide with centre of buoyancy
height immersed
$=2Y_{cm}=2\times \frac{9L}{28}=\frac{9L}{14}$
$\Rightarrow x=9$