An ice block floats in a liquid whose density is less than water. $A$ part of the block is outside the liquid. When the whole of the ice has melted,the liquid level will

$A$ pendulum has a time period $T$ in air. When it is made to oscillate in water,it acquires a time period $T' = \sqrt{2} T$. The density of the pendulum bob is (density of water $= 1$)

$A$ body when fully immersed in a liquid of specific gravity $1.2$ weighs $44 \text{ gwt}$. The same body when fully immersed in water weighs $50 \text{ gwt}$. The mass of the body is (in $\text{ g}$)

Construction of submarines is based on

$A$ 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.]

Two solids $P$ and $Q$ float in water. It is observed that $P$ floats with half of its volume immersed and $Q$ floats with $\frac{2}{3}$ of its volume immersed. The ratio of densities of $P$ and $Q$ is