$A$ 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:

$A$ vertical $U-$tube of uniform inner cross-section contains mercury in both sides of its arms. $A$ glycerin (density = $1.3 \text{ g/cm}^3$) column of length $10 \text{ cm}$ is introduced into one of its arms. Oil of density $0.8 \text{ g/cm}^3$ is poured into the other arm until the upper surfaces of the oil and glycerin are in the same horizontal level. Find the length of the oil column in $\text{cm}$. (Density of mercury = $13.6 \text{ g/cm}^3$)

On opposite sides of a wide vertical vessel filled with water (density $\rho$), two identical holes are drilled, each having cross-section area $A$. The height difference between holes is $x$. The resultant force of reaction of water flowing out of the vessel is ($g$ = acceleration due to gravity).

$A$ 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).

$A$ light semi-cylindrical gate of radius $R$ and length $l$ is pivoted at its midpoint $O$ of the diameter as shown in the figure,holding liquid of density $\rho$. The force $F$ required to prevent the rotation of the gate is equal to:

As shown in the figure, a liquid is at the same level in both arms of a $U$-tube of uniform cross-section when at rest. If the $U$-tube moves with an acceleration '$f$' towards the right, the difference between the liquid heights in the two arms of the $U$-tube will be (acceleration due to gravity $= g$):