$A$ body of density $\rho$ is dropped from rest from a height $h$ into a lake of density $\sigma$,where $\sigma > \rho$. Neglecting all dissipative forces,the maximum depth to which the body sinks before returning to float on the surface is:

$A$ sphere of material of relative density $8$ has a concentric spherical cavity and just sinks in water. If the radius of the sphere is $2 \ cm$,then the volume of the cavity is

$A$ cylindrical block floats vertically in a liquid of density $\rho_{1}$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x_{1}$. Then some amount of another immiscible liquid of density $\rho_{2} (\rho_{2} < \rho_{1})$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x_{2}$ being the fraction of volume of the cylinder inside the liquid of density $\rho_{1}$. The ratio $\rho_{1} / \rho_{2}$ will be

$A$ small metal sphere of density $\varrho$ is dropped from height $h$ into a jar containing liquid of density $\sigma$ $(\sigma > \varrho)$. The maximum depth up to which the sphere sinks is (Neglect damping forces).

The weight of a body in water is one-third of its weight in air. The density of the body is ....... $g/cm^3$. (in $.5$)

$A$ $0.5\ kg$ mass of lead is submerged in a container filled to the brim with water and a block of wood floats on top. The lead mass is slowly lifted from the container by a thin wire and as it emerges into air the level of the water in the container drops a bit. The lead mass is now placed on the block of wood. As the lead is placed on the wood,what happens to the water level in the container?