$A$ 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 is filled with a liquid up to height $H$. $A$ small hole is made in the vessel at a distance $y$ below the liquid surface as shown in the figure. The liquid emerging from the hole strikes the ground at a distance $x$.

$A$ tank filled with fresh water has a hole in its bottom and water is flowing out of it. If the size of the hole is increased,then

$A$ 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

There are two holes, one each along the opposite sides of a wide rectangular tank. The cross-section of each hole is $0.01 \,m^2$ and the vertical distance between the holes is $1 \,m$. The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 \,kg/m^3$, $g = 10 \,m/s^2$)

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