$A$ 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:

$A$ 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

$A$ 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 zero is

$A$ cylindrical vessel open at the top is $20 \ cm$ high and $10 \ cm$ in diameter. $A$ circular hole whose cross-sectional area is $1 \ cm^2$ is cut at the centre of the bottom of the vessel. Water flows from a tube above it into the vessel at the rate of $100 \ cm^3 s^{-1}$. The height of water in the vessel under steady state is ....... $cm$ (Take $g = 1000 \ cm s^{-2}$)

The figure shows two holes in a wide tank containing a liquid column. The water streams coming out of these holes strike the ground at the same point. The height of the liquid column in the tank is $....\ cm$.

$A$ cylindrical vessel filled with water up to the height $H$ becomes empty in time $t_0$ due to a small hole at the bottom of the vessel. If water is filled to a height $4H$,it will flow out in time: