$A$ body having volume $V$ and density $\rho$ is attached to the bottom of a container as shown. The density of the liquid is $d$ (where $d > \rho$). The container has a constant upward acceleration $a$. The tension in the string is:

$A$ cubical block of wood with a side length of $10 \ cm$ floats at the interface between oil and water,with its lower surface horizontal and $4 \ cm$ below the interface. The density of oil is $0.6 \ g/cm^3$. The mass of the block is ......... $gm$.

The reading of a spring balance when a block is suspended from it in air is $60 \, N$. This reading is changed to $40 \, N$ when the block is submerged in water. The specific gravity of the block must be therefore ............

$A$ cubical block of wood of edge $10$ $cm$ and mass $0.92$ $kg$ floats in a tank of water with a layer of oil of relative density $0.6$ to a depth of $4$ $cm$ above the water. When the block attains equilibrium with four of its side edges vertical,which of the following is true?

$A$ 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.]

State Archimedes' principle.