When an air bubble of radius $r$ rises from the bottom to the surface of a lake,its radius becomes $5r/4$ (the pressure of the atmosphere is equal to the $10 \, m$ height of water column). If the temperature is constant and the surface tension is neglected,the depth of the lake is .... $m$ (in $.53$)

$A$ bottle of soda water is grasped by the neck and swung briskly in a vertical circle. Near which portion of the bottle do the bubbles collect?

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

$A$ cylindrical tube,with its base as shown in the figure,is filled with water. It is moving down with a constant acceleration $a$ along a fixed inclined plane with angle $\theta=45^{\circ}$. $P_1$ and $P_2$ are pressures at points $1$ and $2$,respectively,located at the base of the tube. Let $\beta=(P_1-P_2) / (\rho g d)$,where $\rho$ is the density of water,$d$ is the inner diameter of the tube,and $g$ is the acceleration due to gravity. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\beta=0$ when $a=g / \sqrt{2}$
$(B)$ $\beta>0$ when $a=g / \sqrt{2}$
$(C)$ $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a=g / 2$
$(D)$ $\beta=\frac{1}{\sqrt{2}}$ when $a=g / 2$

$A$ cylindrical capillary tube of $0.2 \ mm$ radius is made by joining two capillaries $T_1$ and $T_2$ of different materials having water contact angles of $0^{\circ}$ and $60^{\circ}$,respectively. The capillary tube is dipped vertically in water in two different configurations,case $I$ and $II$ as shown in the figure. Which of the following option$(s)$ is(are) correct?
(Surface tension of water $= 0.075 \ N/m$,density of water $= 1000 \ kg/m^3$,take $g = 10 \ m/s^2$)
$(1)$ The correction in the height of the water column raised in the tube,due to the weight of water contained in the meniscus,will be different for both cases.
$(2)$ For case $I$,if the capillary joint is $5 \ cm$ above the water surface,the height of the water column raised in the tube will be more than $8.75 \ cm$. (Neglect the weight of the water in the meniscus)
$(3)$ For case $I$,if the joint is kept at $8 \ cm$ above the water surface,the height of the water column in the tube will be $7.5 \ cm$. (Neglect the weight of the water in the meniscus)
$(4)$ For case $II$,if the capillary joint is $5 \ cm$ above the water surface,the height of the water column raised in the tube will be $3.75 \ cm$. (Neglect the weight of the water in the meniscus)

The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. $A$ graph showing the speed $v$ with which the liquid is crossing points at a distance $X$ from $O$ along a radius $XO$ would look like: