$A$ wooden block with a coin placed on its top floats in water as shown in the figure. $l$ and $h$ are as shown. After some time,the coin falls into the water. Then:

$A$ 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:

$A$ 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.]

$A$ weightless bag is filled with $5 \,kg$ of water and is weighed in water using a spring balance. The reading of the spring balance is (in $\,kg$)

$A$ block of steel of size $5 \times 5 \times 5 \, cm^3$ is weighed in water. If the relative density of steel is $7$,its apparent weight is ........... $g wt$.

Write the equation for the volume of the partially submerged part of a body floating on the surface of a liquid.