The height of water in a tank is $H$. The range of the liquid emerging out from a hole in the wall of the tank at a depth $\frac{3H}{4}$ from the upper surface of water will be:

There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \text{ atm}$ $(1 \text{ atm} = 10^5 \text{ N/m}^2)$,then the velocity of water flowing from the hole is:

$A$ water tank kept on the ground has an orifice of $2 \,mm$ diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent (in $\,cm$)? (Assume, $g=10 \,m/s^2, \rho_{\text{water}}=10^3 \,kg/m^3$, viscosity $=1$ centi-poise)

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

$A$ 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:

$A$ cylindrical tank having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)