$A$ 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 $cm$)

There are two identical small holes on the opposite sides of a tank filled with a liquid. The tank is open at the top. The difference in height between the two holes is $h$. As the liquid comes out of the two holes,the tank will experience a net horizontal force proportional to

$A$ tank with vertical walls is mounted so that its base is at a height $H$ above the horizontal ground. The tank is filled with water to a depth $h$. $A$ hole is punched in the side wall of the tank at a depth $x$ below the water surface. To have maximum range of the emerging stream,the value of $x$ is

Water is filled in a tank up to a height of $3 \,m$. The base of the tank is at a height of $1 \,m$ above the ground. What should be the height of a hole made in the side of the tank so that water can be sprayed up to the maximum horizontal distance on the ground?

The bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$,it flows out in $t$ seconds. If water is filled to a height $4h$,it will flow out in time:

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: