$A$ block of iron contains a hollow cavity as shown below. The block weighs $6000 \,N$ in air and $4000 \,N$ in water. If the densities of iron and water are $6 \,g/cm^3$ and $1 \,g/cm^3$ respectively, then the volume of the cavity is (assume $g = 10 \,m/s^2$): (in $\,m^3$)

An air bubble of radius $1\,cm$ in water has an upward acceleration of $9.8\,cm\,s^{-2}$. The density of water is $1\,g\,cm^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $.......g$.
$(g = 980\,cm\,s^{-2})$

$A$ small metal sphere of density $\varrho$ is dropped from height $h$ into a jar containing liquid of density $\sigma$ $(\sigma > \varrho)$. The maximum depth up to which the sphere sinks is (Neglect damping forces).

$A$ 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.]

$A$ small wooden ball of density $\rho$ is immersed in water of density $\sigma$ to a depth $h$ and then released. The height $H$ above the surface of water up to which the ball will jump out of water is

$A$ certain block weighs $15 \ N$ in air. It weighs $12 \ N$ when immersed in water. When immersed in another liquid,it weighs $13 \ N$. The relative density of the block is: