$A$ 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:

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 $; m$)?

$A$ cylindrical tank is filled with water to a level of $3 \,m$. $A$ hole is opened at a height of $52.5 \,cm$ from the bottom. The ratio of the area of the hole to that of the cross-sectional area of the tank is $0.1$. The square of the speed with which water will be coming out from the orifice is $(g=10 \,ms^{-2})$.

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 up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

$A$ tank of height $5\, m$ is full of water. There is a hole of cross-sectional area $1\, cm^2$ in its bottom. The initial volume of water that will come out from this hole per second is