$A$ large open-top water tank is completely filled with water. $A$ small hole of diameter $4 \,mm$ is made $10 \,m$ below the water level. The flow rate of water through the hole is (Acceleration due to gravity $= 10 \,m/s^2$)

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

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:

$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water,the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

$A$ vessel completely filled with water has holes $A$ and $B$ at depths $h$ and $3h$ from the top respectively. The hole $A$ is a square of side $L$ and $B$ is a circle of radius $r$. The water flowing out per second from both the holes is same. Then $L$ is equal to

$A$ cylindrical tank has a hole of $1 \, cm^2$ in its bottom. If water is allowed to flow into the tank from a tube above it at the rate of $70 \, cm^3/sec$, then the maximum height up to which water can rise in the tank is .......... $cm$.