Water flows into a cylindrical vessel of large cross-sectional area at a rate of $10^{-4} \; m^3/s$. It flows out from a hole of area $10^{-4} \; m^2$, which has been punched through the base. How high does the water rise in the vessel (in $; m$)?

$A$ liquid is filled in a tank up to a height of $5 \,m$ and the bottom of the tank is at $5 \,m$ height from the ground. $A$ hole is punched in the tank so that the liquid coming out touches the ground at maximum horizontal distance. Then, the maximum horizontal distance is (in $\,m$)

$A$ tank with vertical walls is mounted so that its base is at a height $H$ above the horizontal ground. The tank is filled with water to a depth $h$. $A$ hole is punched in the side wall of the tank at a depth $x$ below the water surface. To have maximum range of the emerging stream,the value of $x$ is

$A$ tank is filled up to a height $2H$ with a liquid and is placed on a platform of height $H$ from the ground. The distance $x$ from the ground where a small hole is punched to get the maximum range $R$ is:

$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $h$ to $h/2$ and from $h/2$ to $0$ is

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).