The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be $75 \ cm$ of $Hg$ and the density of water to be $1/10$ of the density of mercury,the depth of the lake is ....... $m$.

Vapour is injected at a uniform rate into a closed vessel which was initially evacuated. The pressure in the vessel:

The radius of an air bubble at the bottom of a lake is $r$,and it becomes $2r$ when the air bubble rises to the top surface of the lake. If $P \text{ cm}$ of water is the atmospheric pressure,then the depth of the lake is: (in $P$)

The relative humidity on a day,when the partial pressure of water vapour is $0.012 \times 10^5 \, Pa$ at $12^{\circ}C$,is ......... $\%$. (Take the vapour pressure of water at this temperature as $0.016 \times 10^5 \, Pa$.)

Statement $(A)$: When the temperature increases,the viscosity of gases increases and the viscosity of liquids decreases.
Statement $(B)$: Water does not wet an oily glass because the cohesive force of oil is less than that of water.
Statement $(C)$: $A$ liquid will wet a surface of a solid if the angle of contact is greater than $90^{\circ}$.

$(a)$ 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 $\rho_{He} = 0.18 \; kg \, m^{-3}$]