$A$ solid sphere of radius $r$ is floating at the interface of two immiscible liquids of densities $\rho_1$ and $\rho_2$ $(\rho_2 > \rho_1)$,with half of its volume lying in each. The height of the upper liquid column from the interface of the two liquids is $h$. The force exerted on the sphere by the upper liquid is (atmospheric pressure $= p_0$ and acceleration due to gravity is $g$).

$A$ block of wood is floating on water at $0^{\circ} C$ with volume $V_0$ above water. When the temperature of water increases from $0^{\circ} C$ to $10^{\circ} C$,the change in the volume of the block that is above water is best described schematically by which graph?

$A$ candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D$ $(D >> d)$ as shown in the figure. If it is burning at the rate of $2 \text{ cm/hour}$,then the top of the candle will:

$A$ cube having a side of $10 \ cm$ with unknown mass $m$ and a $200 \ g$ mass were hung at two ends of a uniform rigid rod of $27 \ cm$ length. The rod along with the masses was placed on a wedge,keeping the distance between the wedge point and the $200 \ g$ weight as $25 \ cm$. Initially,the masses were not in balance. $A$ beaker is placed beneath the unknown mass and water is added slowly to it. At a given point,the masses were in balance and half the volume of the unknown mass was submerged in the water. (Take the density of the unknown mass to be greater than that of water,the mass did not absorb water,and the water density is $1 \ g/cm^3$). The unknown mass $m$ is . . . . . . $kg$.

$A$ 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.]

$A$ wooden block is floating on water kept in a beaker. $40 \%$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$. The block will then