An object of mass $10 \ kg$ is released from rest in a liquid. If the object moves a distance of $2 \ m$ while sinking in a time duration of $1 \ s$,then the mass of the liquid displaced by the submerged object is $(g = 10 \ m/s^2)$. (in $kg$)

If $W$ is the weight of a body of density $\rho$ in vacuum,then its apparent weight in air of density $\sigma$ is:

$A$ log of wood of mass $120 \ kg$ floats in water. The mass that can be put on the raft to make it just sink should be ....... $kg$ (density of wood = $600 \ kg/m^3$,density of water = $1000 \ kg/m^3$).

$A$ 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

$A$ rectangular block of size $10 \,cm \times 10 \,cm \times 15 \,cm$ is floating in water with the $10 \,cm$ side vertical. If it floats with the $15 \,cm$ side vertical,then the level of water will ..........

$A$ 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.]