$A$ water barrel stands on a table of height $h$. If a small hole is punched in the side of the barrel at its base,it is found that the resultant stream of water strikes the ground at a horizontal distance $R$ from the barrel. The depth of water in the barrel is

$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

$A$ cylindrical vessel of $40 \text{ cm}$ radius is completely filled with water and its capacity is $528 \text{ dm}^3$. The vessel is placed on a solid block of exactly the same height as the vessel. If a small hole is made at $70 \text{ cm}$ below the top of the water level,then the horizontal range of water falling on the ground in the beginning is . . . . . . $\text{cm}$.

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

$A$ large open-top water tank is completely filled with water. $A$ small hole of diameter $4 \,mm$ is made $10 \,m$ below the water level. The flow rate of water through the hole is (Acceleration due to gravity $= 10 \,m/s^2$)

There is a hole of area $A_0$ at the bottom of a cylindrical vessel of cross-sectional area $A$. Water is filled up to a height $h$ and the water flows out in time $t$. If water is filled to a height $4h$,it will flow out in time equal to: