$A$ liquid drop of density $Q$ is floating half-immersed in a liquid of density $d$. What is the diameter of the liquid drop? ($Q$ > $d$, $g = $ acceleration due to gravity, $T = $ surface tension)

$A$ cylindrical furnace has height $(H)$ and diameter $(D)$ both $1 \ m$. It is maintained at temperature $T=360 \ K$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T=360 \ K$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d=0.1 \ m$ and height $h=9 \ m$ above the furnace and exits the chimney. As a result,atmospheric air of density $\rho_a=1.2 \ kg \ m^{-3}$,pressure $P_a$ and temperature $T_a=300 \ K$ enters the furnace. Assume air as an ideal gas,neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects. [Given: The acceleration due to gravity $g=10 \ ms^{-2}$ and $\pi=3.14$]
$(1)$ Considering the air flow to be streamline,the steady mass flow rate of air exiting the chimney is . . . . .$g \ s^{-1}$.
$(2)$ When the chimney is closed using a cap at the top,a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air,due to the stoppage of air flow,are negligible then the value of $\Delta P$ is . . . . .$N \ m^{-2}$.

$A$ table tennis ball has radius $(3 / 2) \times 10^{-2} \text{ m}$ and mass $(22 / 7) \times 10^{-3} \text{ kg}$. It is slowly pushed down into a swimming pool to a depth of $d = 0.7 \text{ m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$,without getting wet,and rises up to a height $H$. Which of the following option$(s)$ is (are) correct?
[Given: $\pi = 22 / 7, g = 10 \text{ ms}^{-2}$,density of water $= 1 \times 10^3 \text{ kg m}^{-3}$,viscosity of water $= 1 \times 10^{-3} \text{ Pa-s}$.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 \text{ J}$.
$(B)$ If we neglect the viscous force in water,then the speed $v = 7 \text{ m/s}$.
$(C)$ If we neglect the viscous force in water,then the height $H = 1.4 \text{ m}$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density $\rho$. The height of the liquid in one vessel is ${h_1}$ and that in the other vessel is ${h_2}$. The area of either base is $A$. The work done by gravity in equalizing the levels when the two vessels are connected is:

An air bubble rises from the bottom of a water tank of height $5 \ m$. If the initial volume of the bubble is $3 \ mm^3$, what will be its volume as it reaches the surface (in $mm^3$)? Assume that its temperature does not change. $[g=9.8 \ m \ s^{-2}, 1 \ atm=10^5 \ Pa, \text{density of water}=1 \ g/cm^3]$

$A$ cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The diameter of the vessel is $10 \, cm$ and the angular speed of rotation is $\omega \, rad \, s^{-1}$. The difference in the height,$h$ (in $cm$),of the liquid at the centre of the vessel and at the side will be