$A$ cylindrical tank of height $0.4\,m$ is open at the top and has a diameter $0.16\,m$. Water is filled in it up to a height of $0.16\,m$. How long will it take to empty the tank through a hole of radius $5 \times 10^{-3}\,m$ in its bottom? (in seconds)

The height of the water level in a tank of uniform cross-section is $5 \,m$. The volume of water leaked in $5 \,s$ through a hole of area $2.4 \,mm^2$ made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity $= 10 \,ms^{-2}$).

$A$ 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:

$A$ water barrel having water up to a depth $d$ is placed on a table of height $h$. $A$ small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance $R$ from the barrel, then the value of $d$ is

$A$ cylindrical tank has a hole of area $2 \,cm^2$ at its bottom. If water is poured into the tank from a tube above it at the rate of $100 \,cm^3 \,s^{-1}$, then the maximum height up to which water can rise in the tank is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $cm$)

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 the hole in $A$. If the liquid mass flux through each hole is equal,then the ratio of the densities of the liquids,$\rho_A / \rho_B$,is: