The three water-filled tanks shown have the same volume and height. If small identical holes are punched near their bottoms,which one will be the first to get empty?

There are two holes, one each along the opposite sides of a wide rectangular tank. The cross-section of each hole is $0.01 \,m^2$ and the vertical distance between the holes is $1 \,m$. The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 \,kg/m^3$, $g = 10 \,m/s^2$)

Water is filled in a container up to a height of $3\,m$. $A$ small hole of area $A_0$ is punched in the wall of the container at a height of $52.5\,cm$ from the bottom. The cross-sectional area of the container is $A$. If $A_0/A = 0.1$,then $v^2$ is......... $m^2/s^2$ (where $v$ is the velocity of water coming out of the hole).

$A$ 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:

$A$ cylinder containing water up to a height of $25 \ cm$ has a hole of cross-section $\frac{1}{4} \ cm^2$ in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out?