There is a hole of area $A_0$ at the bottom of a cylindrical vessel of cross-sectional area $A$. Water is filled up to a height $h$ and the water flows out in time $t$. If water is filled to a height $4h$,it will flow out in time equal to:

In a cylindrical vessel containing liquid of density $\rho$,there are two holes in the side walls at heights of $h_1$ and $h_2$ respectively,such that the range of efflux at the bottom of the vessel is the same. The height of a hole,for which the range of efflux would be maximum,is:

There are two identical small holes of cross-sectional area $a$ on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. The tank is resting on a smooth horizontal surface. The horizontal force that must be applied to the tank to keep it in equilibrium 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

Tanks $A$ and $B$ open at the top contain two different liquids up to a certain height. $A$ hole is made in the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$. If the liquid mass flux through each hole is equal,then the ratio of the densities of the liquids respectively,is

$A$ tank is filled with water up to a height $H$. Water is allowed to come out of a hole in one of the walls at a depth $D$ below the surface of water. Express the horizontal distance $x$ in terms of $H$ and $D$.