$A$ 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 $0$ is

$A$ vessel completely filled with water has two holes $P$ and $Q$ at depths $2h$ and $8h$ from the top,respectively. Hole $P$ is a square of side $a$ and hole $Q$ is a circle of radius $r$. If the volume of water flowing out per second from both holes is the same,then the side $a$ of hole $P$ 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

$A$ small hole of area of cross-section $2 \; mm^{2}$ is present near the bottom of a fully filled open tank of height $2 \; m$. Taking $g = 10 \; m/s^{2}$,the rate of flow of water through the open hole would be nearly ......... $\times 10^{-6} \; m^{3}/s$.

$A$ 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 $1 \ cm^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \ cm^3/s$,then the maximum height up to which water can rise in the tank is: $(g = 9.8 \ m/s^2)$ (in $cm$)