$A$ tube is attached as shown in a closed vessel containing water. The velocity of water coming out from a small hole is:

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:

$A$ tank is filled with water to a height $H$. $A$ hole is made in one of the walls at a depth $D$ below the water surface. The distance $x$ from the foot of the wall at which the stream of water coming out of the tank strikes the ground is given by .............

$A$ 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 $cm$)

$A$ 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 $\,cm$)

$A$ vessel of area of cross-section $A$ has liquid to a height $H$. There is a hole at the bottom of the vessel having area of cross-section $a$. The time taken to decrease the level from $H_1$ to $H_2$ will be: