There is a hole at the bottom of a water tank. The total pressure at the bottom is $3 \, atm$ $(1 \, atm = 10^{5} \, N/m^{2})$. What is the velocity of the water coming out of the hole?

$A$ 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 $cm$)

The top of a water tank is open to air and its water level is maintained. It is discharging $0.74\,m^3$ of water per minute through a circular opening of $2\,cm$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to........ $m$

In a cylindrical water tank,there are two small holes $A$ and $B$ on the wall. Hole $A$ is at a depth of $h_1$ from the water surface,and hole $B$ is at a height of $h_2$ from the bottom of the tank. The total height of the water surface from the bottom of the tank is $H$. Water coming out from both holes strikes the ground at the same point $S$. Find the ratio of $h_1$ and $h_2$.

$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

$A$ 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 $cm$)