$A$ 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

$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

$A$ square hole of side length $l$ is made at a depth of $h$ and a circular hole of radius $r$ is made at a depth of $4h$ from the surface of water in a water tank kept on a horizontal surface. If $l << h$ and $r << h$,and the rate of water flow from the holes is the same,then $r$ is equal to:

$A$ large open tank containing water has two holes in its wall. $A$ square hole of side $a$ is made at a depth $y$ and a circular hole of radius $r$ is made at a depth $16y$ from the surface of the water. If an equal amount of water comes out through both the holes per second,then the relation between $r$ and $a$ will be:

$A$ 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 $= 10 \,m/s^2$)

$A$ 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)