$A$ cylindrical tank with a large diameter is filled with water. Water drains out through a hole at the bottom of the tank. If the cross-sectional area of the hole is $6 \ cm^2$,then the drainage rate in $m^3 s^{-1}$ when the depth of the water is $0.2 \ m$ is:

$A$ cylinder of height $h$ is filled with water and is kept on a block of height $h/2$. The level of water in the cylinder is kept constant. Four holes numbered $1, 2, 3$ and $4$ are at the side of the cylinder and at heights $0, h/4, h/2$ and $3h/4$ from the base of the cylinder, respectively. When all four holes are opened together, the hole from which water will reach the farthest distance on the plane $PQ$ is the hole number:

There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \text{ atm}$ $(1 \text{ atm} = 10^5 \text{ N/m}^2)$,then the velocity of water flowing from the hole is:

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank is:

$A$ cylindrical tank has a hole of $1 \ cm^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \ cm^3/s$,then the maximum height up to which water can rise in the tank is: $(g = 9.8 \ m/s^2)$ (in $cm$)

There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \ atm$,then the velocity of water flowing from the hole is $(1 \ atm = 10^5 \ Pa)$ :-