In a turbulent flow,the velocity of the liquid molecules in contact with the walls of the tube is

Two identical cylindrical vessels with their bases at the same level, each contains a liquid of density $d$. 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

$A$ 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)

$A$ cylindrical tank of height $1 \ m$ and cross-sectional area $A = 4000 \ cm^2$ is initially empty. It is kept under a tap of cross-sectional area $a_1 = 1 \ cm^2$. Water starts flowing from the tap at $t = 0$ with a speed $v_1 = 2 \ m/s$. There is a small hole in the base of the tank of cross-sectional area $a_2 = 0.5 \ cm^2$. The variation of the height of water in the tank (in meters) with time $t$ is best depicted by:

$A$ drum of radius $R$ full of liquid of density $d$ is rotated at angular velocity $\omega \ rad/s$. The increase in pressure at the centre of the drum will be

$A$ liquid kept in a cylindrical vessel is rotated about a vertical axis passing through the center of the circular base. The difference in the heights of the liquid at the center of the vessel and its edge is ($R=$ radius of vessel,$\omega=$ angular velocity of rotation,$g=$ acceleration due to gravity).