$A$ block of volume $V$ and density $\sigma_b$ is placed in a liquid of density $\sigma_l$ (where $\sigma_l > \sigma_b$). If the block is moved upward by a height $h$ while remaining completely submerged in the liquid,what is the increase in the gravitational potential energy of the system?

$A$ hollow sphere of volume $V$ is floating on the water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water?

$A$ hollow spherical body of outer and inner radii of $4 \ cm$ and $2 \ cm$ respectively floats half submerged in a liquid of density $2.0 \ g \ cm^{-3}$. The density of the material of the sphere is (in $g \ cm^{-3}$)

$A$ 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.]

$A$ hollow sphere of external radius $R$ and thickness $t$ $(t \ll R)$ is made of a metal of density $\rho$. The sphere will float in water if:

$A$ wooden block,with a coin placed on its top,floats in water as shown in the figure. The distances $l$ and $h$ are shown. After some time,the coin falls into the water. Then: