$A$ 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

The height of water in a tank is $H$. The range of the liquid emerging out from a hole in the wall of the tank at a depth $\frac{3H}{4}$ from the upper surface of water will be:

$A$ container of large uniform cross-sectional area $A$ is resting on a horizontal surface and holds two immiscible,non-viscous,and incompressible liquids of densities $d$ and $2d$,each of height $\frac{H}{2}$,as shown. The upper surface of the liquid with lower density is open to the atmosphere. $A$ small hole is made on the wall of the container at height $h$ $(h < \frac{H}{2})$. The initial speed of efflux of the liquid at the hole is:

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 $; m$)?

Equal volumes of two immiscible liquids of densities $\rho$ and $2\rho$ are filled in a vessel as shown in the figure. Two small holes are punched at depths $h/2$ and $3h/2$ from the surface of the lighter liquid. If $v_1$ and $v_2$ are the velocities of efflux at these two holes,then $v_1/v_2$ is:

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: