$A$ ball whose density is $0.4 \times 10^3 \ kg/m^3$ falls into water from a height of $9 \ cm$. To what depth does the ball sink in $cm$?

$A$ container of large surface area is filled with a liquid of density $\rho$. $A$ cubical block of side edge $a$ and mass $M$ is floating in it with four-fifths of its volume submerged. If a coin of mass $m$ is placed gently on the top surface of the block,it is just submerged. $M$ is:

$A$ small wooden ball of density $\rho$ is immersed in water of density $\sigma$ to a depth $h$ and then released. The height $H$ above the surface of water up to which the ball will jump out of water is

$A$ hemispherical portion of radius $R$ is removed from the bottom of a cylinder of radius $R$. The volume of the remaining cylinder is $V$ and its mass is $M$. It is suspended by a string in a liquid of density $\rho$ where it stays vertical. The upper surface of the cylinder is at a depth $h$ below the liquid surface. The force on the bottom of the liquid is

$A$ homogeneous solid cylinder of length $L$ $(L < H/2)$, cross-sectional area $A$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L/4$ in the denser liquid (density $2d$) and $3L/4$ in the lighter liquid (density $d$). The lower density liquid is open to the atmosphere having pressure $P_0$. Then, the density $D$ of the solid is given by:

$A$ wooden cube just floats inside water with a $200 \,g$ mass placed on it. When the mass is removed,the cube floats with its top surface $2 \,cm$ above the water level. The side of the cube is ......... $cm$.