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

$A$ spring balance $A$ reads $2 \,kg$ with a block of mass $m$ suspended from it. Another balance $B$ reads $3 \,kg$ when a beaker with a liquid is put on its pan. The two balances are now so arranged that the hanging mass $m$ is fully immersed inside the liquid in the beaker as shown in the figure. In this situation,

$A$ tub is filled with water and a wooden cube $10 \ \text{cm} \times 10 \ \text{cm} \times 10 \ \text{cm}$ is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87 \ \text{cm}$. The mass of the metal coin is . . . . . . gram. (Take water density as $1 \ \text{g/cm}^3$ and density of wood as $0.4 \ \text{g/cm}^3$)

$A$ metallic body of material with density of $8000\ kg/m^3$ has a cavity inside. $A$ spring balance shows its mass to be $10.0\ kg$ in air and $7.5\ kg$ when immersed in water. The ratio of the volume of the cavity to the volume of the material of the body must be

$A$ solid sphere of density $\rho_s$ which is $\eta$ times lighter than water (density $\rho_w$) is suspended in a water tank by a string tied to its base as shown in the figure. If the mass of the sphere is $m$,then the tension in the string is given by:

$A$ spring balance reads $200 \,gF$ when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine solution,what will be the new reading of the spring balance? The specific gravity of lead and brine are $11.4$ and $1.1$ respectively. (Result in $gF$)