$A$ 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:

$A$ $0.5 \,kg$ block of brass (density $=8 \times 10^3 \,kg \,m^{-3}$) is suspended from a string. What is the tension in the string if the block is completely immersed in water? $(g=10 \,ms^{-2})$

An air bubble of radius $1\,cm$ in water has an upward acceleration of $9.8\,cm\,s^{-2}$. The density of water is $1\,g\,cm^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $.......g$.
$(g = 980\,cm\,s^{-2})$

An iceberg floats in water with part of it submerged. What is the fraction of the volume of the iceberg submerged if the density of ice is ${\rho _i} = 0.917 \, g/cm^3$?

$A$ piece of steel has a weight $W$ in air,$W_1$ when completely immersed in water,and $W_2$ when completely immersed in an unknown liquid. The relative density (specific gravity) of the liquid is:

$A$ gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. $A$ gas column under gravity,for example,does not have uniform density (and pressure). As you might expect,its density decreases with height. The precise dependence is given by the so-called law of atmospheres:
$n_{2}=n_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
where $n_{2}, n_{1}$ refer to number density at heights $h_{2}$ and $h_{1}$ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
$n_{2}=n_{1} \exp \left[-m g N_{A}\left(\rho-\rho^{\prime}\right)\left(h_{2}-h_{1}\right) /(\rho R T)\right]$
where $\rho$ is the density of the suspended particle,and $\rho^{\prime}$ that of the surrounding medium. [$N_{A}$ is Avogadro's number,and $R$ the universal gas constant.]