$A$ 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,

$A$ vessel contains oil (density = $0.8 \; g/cm^3$) over mercury (density = $13.6 \; g/cm^3$). $A$ homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g/cm^3$ is:

$A$ sphere of material of relative density $8$ has a concentric spherical cavity and just sinks in water. If the radius of the sphere is $2 \ cm$,then the volume of the cavity is

$A$ wooden block floating in a bucket of water has $\frac{4}{5}$ of its volume submerged. When a certain amount of oil is poured into the bucket,it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water 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.]

$A$ body is suspended by a light string. The tensions in the string when the body is in air,when the body is totally immersed in water,and when the body is totally immersed in a liquid are respectively $40.2 \,N$,$28.4 \,N$,and $16.6 \,N$. The density of the liquid is