When an air bubble rises from the bottom of a lake to the surface,its radius is doubled. The atmospheric pressure is equal to that of a column of water of height $H$. The depth of the lake is:

Two solid spheres $A$ and $B$ of equal volumes but of different densities $d_A$ and $d_B$ are connected by a string. They are fully immersed in a fluid of density $d_F$. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if:
$(A)$ $d_A < d_F$
$(B)$ $d_B > d_F$
$(C)$ $d_A + d_B = 2d_F$
$(D)$ $d_A > d_F$

$A$ large number of droplets,each of radius $r$,coalesce to form a bigger drop of radius $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. The velocity of the drop is ($T=$ surface tension,$\rho =$ density)

In the figure shown,the heavy cylinder (radius $R$) resting on a smooth surface separates two liquids of densities $2\rho$ and $3\rho$. The height '$h$' for the equilibrium of the cylinder must be

The heart of a man pumps $5 \, \text{litres}$ of blood through the arteries per minute at a pressure of $150 \, \text{mm}$ of mercury. If the density of mercury is $13.6 \times 10^3 \, \text{kg/m}^3$ and $g = 10 \, \text{m/s}^2$, then the power (in $\text{watt}$) is:

$(a)$ It is known that the density $\rho$ of air decreases with height $y$ as $\rho = \rho_{0} e^{-y / y_{0}}$,where $\rho_{0} = 1.25 \; kg \, m^{-3}$ is the density at sea level,and $y_{0}$ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains constant (isothermal conditions). Also,assume that the value of $g$ remains constant.
$(b)$ $A$ large $He$ balloon of volume $1425 \; m^{3}$ is used to lift a payload of $400 \; kg$. Assume that the balloon maintains a constant radius as it rises. How high does it rise?
[Take $y_{0} = 8000 \; m$ and $\rho_{He} = 0.18 \; kg \, m^{-3}$]