$A$ cylindrical vessel filled with water up to the height $H$ becomes empty in time $t_0$ due to a small hole at the bottom of the vessel. If water is filled to a height $4H$,it will flow out in time:

$A$ 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 $= 10 \,m/s^2$)

$A$ 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:

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 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 .............

$A$ rectangular tank is placed on a horizontal ground and is filled with water to a height $H$ above the base. $A$ small hole is made on one vertical side at a depth $D$ below the level of the water in the tank. The distance $x$ from the bottom of the tank at which the water jet from the tank will hit the ground is