$A$ liquid is filled in a tank up to a height of $5 \,m$ and the bottom of the tank is at $5 \,m$ height from the ground. $A$ hole is punched in the tank so that the liquid coming out touches the ground at maximum horizontal distance. Then, the maximum horizontal distance is (in $\,m$)

$A$ cylindrical vessel, open at the top, contains $15 \text{ litres}$ of water. Water drains out through a small opening at the bottom. $5 \text{ litres}$ of water comes out in time $t_1$, the next $5 \text{ litres}$ in further time $t_2$, and the last $5 \text{ litres}$ in further time $t_3$. Then,

$A$ large tank open to the atmosphere at the top and filled with water develops a small hole in the side at a point $20 \,m$ below the water level. If the rate of flow of water from the hole is $3 \times 10^{-3} \,m^3/min$, then the area of the hole is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,mm^2$)

$A$ 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

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank 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 $= 1000 \,kg/m^3$, $g = 10 \,m/s^2$)